Polynomials ⇒⇒ Exercise 2.4

Question 1 (i)

Use suitable identities to find the following products:
(i) (x +4) (x +10)

Solution :

Given , (x +4) (x +10)

Using suitable identity : (x + a) (x + b) = x² + (a + b)x + ab

Here, we have a = 4 and b = 10

Substituting the value of a and b in given expression, we get;

$$ (x + 4)(x + 10) $$

$$⇒ x^2 + (4 + 10) x + 4 × 10 $$

$$⇒ x^2 + 14 x + 40 $$

Question 1 (ii)

Use suitable identities to find the following products:
(ii) (x + 8)(x – 10)

Solution :

Given , (x + 8)(x – 10)

Using suitable identity : (x + a) (x + b) = x² + (a + b)x + ab

Here, we have a = 8 and b = (-10)

Substituting the value of a and b in given expression, we get;

$$ (x + 8)(x + (-10)) $$

$$⇒ x^2 + [8 + (-10)] x + [ 8 × (-10)] $$

$$⇒ x^2 + (-2)x -80 $$

$$⇒ x^2 - 2x - 80 $$

Question 1 (iii)

Use suitable identities to find the following products:
(iii) (x + 8)(x – 10)

Solution :

Given , (3x + 4)(3x – 5)

Using suitable identity : (x + a) (x + b) = x² + (a + b)x + ab

Here, we have x = 3x , a = 4 and b = (-5)

Substituting the value of x, a and b in given expression, we get;

$$ (3x + 4)(3x + (-5)) $$

$$⇒ 3x^2 + [4 + (-5)] 3x + [4 × (-5)] $$

$$⇒ 9x^2 + (-1)3x - 20 $$

$$⇒ 9x^2 - 3x - 20 $$

Question 1 (iv)

Use suitable identities to find the following products:
(iv) (y² + ${3 \over 2} $)(y² – ${3 \over 2} $)

Solution :

Given , (y² + ${3 \over 2} $)(y² – ${3 \over 2} $)

Using suitable identity : (a + b) (a - b) = a² - b²

Here, we have a = y² , b = 3/2

Substituting the value of a and b in given expression, we get;

$$⇒ (y^2)^2 - ({3 \over 2})^2 $$

$$⇒ y^4 - ({9 \over 4})$$

Question 1 (v)

Use suitable identities to find the following products:
(v) (3 – 2x)(3 + 2x)

Solution :

Given , (3 – 2x)(3 + 2x)

Using suitable identity : (a + b) (a - b) = a² - b²

Here, we have a = 3 , b = 2x

Substituting the value of a and b in given expression, we get;

$$⇒ (3)^2 - (2x)^2 $$

$$⇒ 9 - 4x^2 $$

Question 2 (i)

Evaluate the following products without multiplying directly:
(i) 103 × 107

Solution :

Given , 103 × 107

We can write : = ( 100 + 3 ) ( 100 + 7 )

Using the identity : (x + a) (x + b) = x² + (a + b)x + ab

Here, we have x = 100, a = 3 and b = 7

Substituting the value of x, a and b in given expression, we get;

$$ ( 100 + 3 ) ( 100 + 7 ) $$

$$⇒ (100)^2 + [3 + 7]× 100 + [3 × 7] $$

$$⇒ 10000 + (10)× 100 + 21 $$

$$⇒ 10000 + 1000 + 21 $$

$$⇒ 11021 $$

Question 2 (ii)

Evaluate the following products without multiplying directly:
(ii) 95 × 96

Solution :

Given , 95 × 96

We can write : = ( 90 + 5 ) ( 90 + 6 )

Using the identity : (x + a) (x + b) = x² + (a + b)x + ab

Here, we have x = 90, a = 5 and b = 6

Substituting the value of x, a and b in given expression, we get;

$$ ( 90 + 5 ) ( 90 + 6 ) $$

$$⇒ (90)^2 + [5 + 6]× 90 + [5 × 6] $$

$$⇒ 8100 + (11)× 90 + 30 $$

$$⇒ 8100 + 990 + 30 $$

$$⇒ 9120 $$

Question 2 (iii)

Evaluate the following products without multiplying directly:
(iii) 104 × 96

Solution :

Given , 104 × 96

We can write : = ( 100 + 4 ) ( 100 - 4 )

Using the identity : (a + b) (a - b) = a² - b²

Here, we have, a = 100 and b = 4

Substituting the value of x, a and b in given expression, we get;

$$ ( 100 + 4 ) ( 100 - 4 ) $$

$$⇒ (100)^2 - (4)^2 $$

$$⇒ 10000 - 16 $$

$$⇒ 9984 $$

Question 3 (i)

Factorise the following using appropriate identities:
(i) 9x² + 6xy + y²

Solution :

Given , 9x² + 6xy + y²

$$⇒ 3^2x^2 + 6xy +y^2 $$

$$⇒ (3x)^2 + 6xy +(y)^2 $$

$$⇒ (3x)^2 + 2× (3x)(y) +(y)^2 $$

This expression fits the form of the Perfect Square Trinomial (Addition) : a² + 2ab + b² = (a + b)²

so a = 3x and b = y

Therefore, the factorisation is:

$$⇒ (3x + y )^2 $$

$$⇒(3x + y )(3x + y )$$

Question 3 (ii)

Factorise the following using appropriate identities:
(ii) 4y² - 4y + 1

Solution :

Given , 4y² - 4y + 1

$$⇒ 2^2y^2 - 4y + 1^2 $$

$$⇒ (2y)^2 - 4y +(1)^2 $$

$$⇒ (2y)^2 - 2× (2y)(1) +(1)^2 $$

This expression fits the form of the Perfect Square Trinomial (Subtraction) : a² - 2ab + b² = (a - b)²

so a = 2y and b = 1

Therefore, the factorisation is:

$$⇒ (2y - 1 )^2 $$

$$⇒(2y - 1 )(2y - 1 )$$

Question 3 (iii)

Factorise the following using appropriate identities:
(iii) x² - ${y^2 \over 100} $

Solution :

Given , x² - ${y^2 \over 100} $

$$⇒ x^2 - {y^2 \over 10^2} $$

$$⇒ x^2 - ({y \over 10})^2 $$

This expression fits the form of the Difference of Squares : a² - b² = (a + b) (a - b)

so a = x and b = ${y \over 10} $

Therefore, the factorisation is:

$$⇒ (x + {y \over 10})(x - {y \over 10}) $$

Question 4 (i)

Expand each of the following, using suitable identities :
(i) (x + 2y + 4z)²

Solution :

Given , (x + 2y + 4z)²

To expand each of the following expressions, we'll use the algebraic identity for the square of a trinomial:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Let, a = x and b = 2y , and c = 4z

Substituting the value of a and b , c in given expression, we get;

$⇒ (x + 2y + 4z)^2 $

$⇒ (x)^2 + (2y)^2 + (4z)^2 $+$ 2(x)(2y) + 2(2y)(4z) + 2(4z)(x) $

$⇒ x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8zx $

Question 4 (ii)

Expand each of the following, using suitable identities :
(ii) (2x - y + z)²

Solution :

Given , (2x - y + z)²

To expand each of the following expressions, we'll use the algebraic identity for the square of a trinomial:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Let, a = 2x and b = -y , and c = z

Substituting the value of a and b , c in given expression, we get;

$⇒ (2x - y + z)^2 $

$⇒ (2x)^2 + (-y)^2 + (z)^2 $ + $ 2(2x)(-y) + 2(-y)(z) + 2(z)(2x) $

$⇒ 4x^2 + y^2 + z^2 - 4xy - 2yz + 4zx $

Question 4 (iii)

Expand each of the following, using suitable identities :
(iii) (–2x + 3y + 2z)²

Solution :

Given , (–2x + 3y + 2z)²

To expand each of the following expressions, we'll use the algebraic identity for the square of a trinomial:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Let, a = -2x and b = 3y , and c = 2z

Substituting the value of a and b , c in given expression, we get;

$⇒ (–2x + 3y + 2z)^2 $

$⇒ (-2x)^2 + (3y)^2 + (2z)^2 $+$ 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x) $

$⇒ 4x^2 + 9y^2 + 4z^2 - 12xy + 12yz - 8zx $

Question 4 (iv)

Expand each of the following, using suitable identities :
(iv) (3a - 7b -c)²

Solution :

Given , (3a - 7b -c)²

To expand each of the following expressions, we'll use the algebraic identity for the square of a trinomial:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Let, a = 3a and b = -7b , and c = -c

Substituting the value of a and b , c in given expression, we get;

$⇒ (3a -7b -c)^2 $

$⇒ (3a)^2 + (-7b)^2 + (-c)^2 $ +$ 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a) $

$⇒ 9a^2 + 49b^2 + c^2 - 42ab + 14bc - 6ca $

Question 4 (v)

Expand each of the following, using suitable identities :
(v) (– 2x + 5y – 3z)²

Solution :

Given , (– 2x + 5y – 3z)²

To expand each of the following expressions, we'll use the algebraic identity for the square of a trinomial:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Let, a = -2x and b = 5y ,and c = -3z

Substituting the value of a and b , c in given expression, we get;

$⇒ (– 2x + 5y – 3z)^2 $

$⇒ (-2x)^2 + (5y)^2 + (-3z)^2 $ + $ 2(-2x)(5y) + 2(5y)(-3z) + 2(-3z)(-2x) $

$⇒ 4x^2 + 25y^2 + 9z^2 - 20xy - 30yz + 12zx $

Question 4 (vi)

Expand each of the following, using suitable identities :
(vi) (${1 \over 4} $a - ${1 \over 2} $b + 1)²

Solution :

Given , (${1 \over 4} $a - ${1 \over 2} $b + 1)²

To expand each of the following expressions, we'll use the algebraic identity for the square of a trinomial:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Let, a = ${1 \over 4}$a , b = - ${1 \over 2} $b ,and c = 1

Substituting the value of a and b , c in given expression, we get;

$⇒ ({1 \over 4}a -{1 \over 2}b + 1)^2 $

$⇒ ({1 \over 4}a)^2 + (-{1 \over 2}b)^2 + (1)^2 $ $+ 2({1 \over 4}a)(-{1 \over 2}b) + 2(-{1 \over 2}b)(1) + 2(1)({1 \over 4}a) $

$⇒ {1 \over 16}a^2 +{1 \over 4}b^2 + 1 $ $ -{1 \over 4}ab -b + {1 \over 2}a $

Question 5 (i)

Factorise:
(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz

Solution :

Given , 4x² + 9y² + 16z² + 12xy – 24yz – 16xz

This can be re-written as:

$⇒ (2^2x^2) + (3^2y^2) + ( 4^2z^2) $ +$ 12xy – 24yz – 16xz $

$⇒ (2x)^2 + (3y)^2 + (4z)^2 $ + $ 12xy – 24yz – 16xz $

It can be observed that the coefficient value of yz and xz are negative, and the z is common in both term .

Hence ,

$⇒ (2x)^2 + (3y)^2 + (- 4z)^2 $ + $ 12xy – 24yz – 16xz $

$⇒ (2x)^2 + (3y)^2 + (- 4z)^2 $ $ + 2 × (2x) (3y) + 2 ×(3y)(- 4z)$ $+ 2 × (- 4z)(2x) $

Using suitable identity : (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Let, a = 2x , b = 3y , c = -4z

Substituting the value of a and b , c in given expression, we get;

$⇒ (2x + 3y – 4z)^2 $

Thus, the factorization is

$⇒ (2x + 3y – 4z) (2x + 3y – 4z) $

Question 5 (ii)

Factorise:
(ii) 2x² + y² + 8z² – 2√2 xy + 4√2 yz – 8xz.

Solution :

Given , 2x² + y² + 8z² – 2√2 xy + 4√2 yz – 8xz.

This can be re-written as:

$⇒ (√2^2)x^2 + y^2 + (2√2)^2z^2 $ - $ 2√2 xy + 4√2 yz – 8xz $

$⇒ (√2x)^2 + (y)^2 + (2√2z)^2 $ - $ 2√2 xy + 4√2 yz – 8xz $

It can be observed that the coefficient value of xy and xz are negative, and the x is common in both term .

Hence ,

$⇒ (-√2x)^2 + (y)^2 + (2√2z)^2 $ - $ 2√2 xy + 4√2 yz – 8xz $

$⇒ (-√2x)^2 + (y)^2 + (2√2z)^2 $ $+ 2 × (-√2x) (y) + 2 ×(y)(2√2z)$ $+ 2 × (-√2x)(2√2z) $

Let, a = -√2x , b = y , c = 2√2z

Using suitable identity : (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Substituting the value of a and b , c in given expression, we get;

$⇒ (-√2x + y +2√2z)^2 $

Thus, the factorization is

$ (-√2x + y +2√2z)$ $(-√2x + y +2√2z) $

Question 6 (i)

Write the following cubes in expanded form :
(i) (2x + 1)³

Solution :

Given , (2x + 1)³

To expand the given cubes, we will use the binomial expansion identities for the Cube of a sum:

(a + b)³ = a³ + b³ + 3ab (a + b)

Let, a = 2x and b = 1

Substituting the value of a and b in given expression, we get;

$⇒ (2x + 1)^3 $

$⇒ (2x)^3 + (1)^3 $ + $ 3×(2x)(1) ×(2x + 1) $

$⇒ 8x^3 + 1 + 6x ×(2x + 1) $

$⇒ 8x^3 + 1 + 12x^2 + 6x $

$⇒ 8x^3 + 12x^2 + 6x + 1 $

Question 6 (ii)

Write the following cubes in expanded form :
(ii) (2a – 3b)³

Solution :

Given , (2a – 3b)³

To expand the given cubes, we will use the binomial expansion identities for the Cube of a difference:

(x – y)³ = x³ - y³ - 3xy (x - y)

Let, x = 2a and y = 3b

Substituting the value of x and y in given expression, we get;

$⇒ (2a – 3b)^3 $

$⇒ (2a)^3 - (3b)^3 $ - $ 3×(2a)(3b) × (2a -3b) $

$⇒ 8a^3 - 27b^3 - 18ab × (2a -3b) $

$⇒ 8a^3 - 27b^3 - 36a^2b + 54ab^2 $

Question 6 (iii)

Write the following cubes in expanded form :
(iii) (${3 \over 2} $x + 1)³

Solution :

Given , (${3 \over 2} $x + 1)³

To expand the given cubes, we will use the binomial expansion identities for the Cube of a sum:

(a + b)³ = a³ + b³ + 3ab (a + b)

Let, a = ${3 \over 2} $x and b = 1

Substituting the value of a and b in given expression, we get;

$⇒ ({3 \over 2}x + 1)^3 $

$⇒ ({3 \over 2}x)^3 + (1)^3 $ + $ 3×({3 \over 2}x)(1) ×({3 \over 2}x + 1) $

$⇒ {27 \over 8}x^3 + 1 + {9 \over 2}x × ({3 \over 2}x + 1) $

$⇒ {27 \over 8}x^3 + 1 + {27 \over 4}x^2 + {9 \over 2}x $

$⇒ {27 \over 8}x^3 + {27 \over 4}x^2 + {9 \over 2}x + 1 $

Question 6 (iv)

Write the following cubes in expanded form :
(iv) (x - ${2 \over 3}y $)³

Solution :

Given , (x - ${2 \over 3}y $)³

To expand the given cubes, we will use the binomial expansion identities for the Cube of a difference:

(a - b)³ = a³ - b³ - 3ab (a - b)

Let, a = x and b = ${2 \over 3}y $

Substituting the value of a and b in given expression, we get;

$⇒ (x - {2 \over 3}y)^3 $

$⇒ (x)^3 - ({2 \over 3}y)^3 $ - $ 3×(x)({2 \over 3}y) ×(x - {2 \over 3}y) $

$⇒ x^3 - {8 \over 27}y^3 $ - $ 2xy × (x - {2 \over 3}y) $

$⇒ x^3 - {8 \over 27}y^3 $ - $ 2x^2y + {4 \over 3}xy^2 $

$⇒ x^3 - 2x^2y $ - $ {8 \over 27}y^3 + {4 \over 3}xy^2 $

Question 7 (i)

Evaluate the following using suitable identities:
(i) (99)³

Solution :

Given , (99)³

We can write 99 as 100−1.

= ( 100 - 1 )³

We'll use the identity for the cube of a difference : (a - b)³ = a³ - b³ - 3ab (a - b)

Let, a = 100 and b = 1

Substituting the value of a and b in given expression, we get;

$⇒ ( 100 - 1 )^3 $

$⇒ (100)^3 - (1)^3 $ - $ 3×(100)(1) ×(100 - 1) $

$⇒ 1000000 - 1 - 300 × 99 $

$⇒ 1000000 - 1 - 297100 $

$⇒ 970299 $

Question 7 (ii)

Evaluate the following using suitable identities:
(ii) (102)³

Solution :

Given , (102)³

We can write 102 as 100 + 2.

= ( 100 + 2 )³

We'll use the identity for the cube of a sum : (a + b)³ = a³ + b³ + 3ab (a + b)

Let, a = 100 and b = 2

Substituting the value of a and b in given expression, we get;

$⇒ ( 100 + 2 )^3 $

$⇒ (100)^3 + (2)^3 $ + $ 3×(100)(2) ×(100 + 2) $

$⇒ 1000000 + 8 + 600 × 102 $

$⇒ 1000008 + 61200 $

$⇒ 1061208 $

Question 7 (iii)

Evaluate the following using suitable identities:
(iii) (998)³

Solution :

Given , (998)³

We can write 99 as 1000−2.

= ( 1000 - 2 )³

We'll use the identity for the cube of a difference : (a - b)³ = a³ - b³ - 3ab (a - b)

Let, a = 1000 and b = 2

Substituting the value of a and b in given expression, we get;

$⇒ ( 1000 - 2 )^3 $

$⇒ (1000)^3 - (2)^3 $ - $ 3×(1000)(2) ×(1000 - 2) $

$⇒ 1000000000 - 8 - 6000 × 998 $

$⇒ 1000000000 - 8 - 5988000 $

$⇒ 1000000000 - 5988008 $

$⇒ 994011992 $

Question 8 (i)

Factorize each of the following :
(i) 8a³ + b³ + 12a²b + 6ab²

Solution :

Given , 8a³ + b³ + 12a²b + 6ab²

This can be re-written as:

$⇒ (2^3)a^3 + b^3 $ + $ 12a^2b + 6ab^2 $

$⇒ (2a)^3 + (b)^3 $ + $ 12a^2b + 6ab^2 $

$⇒ (2a)^3 + (b)^3 $ + $ 6ab( 2a+ b) $

$⇒ (2a)^3 + (b)^3 $ + $ 3 × 2a × b ( 2a+ b) $

Let, x = 2a , y = b

Using suitable identity : (x + y)³ = x³ + y³ + 3xy (x + y)

Substituting the value of x and y in given expression, we get;

$⇒ (2a +b)^3 $

Thus, the factorization is

$⇒ (2a + b)(2a + b)(2a + b) $

Question 8 (ii)

Factorize each of the following :
(ii) 8a³ - b³ - 12a²b + 6ab²

Solution :

Given , 8a³ - b³ - 12a²b + 6ab²

This can be re-written as:

$⇒ (2^3)a^3 - b^3 $ - $ 12a^2b + 6ab^2 $

$⇒ (2a)^3 - (b)^3 $ - $ 12a^2b + 6ab^2 $

$⇒ (2a)^3 - (b)^3 $ - $ 6ab( 2a - b) $

$⇒ (2a)^3 - (b)^3 $ - $ 3 × 2a × b( 2a - b) $

Let, x = 2a , y = b

Using suitable identity : (x - y)³ = x³ - y³ - 3xy (x - y)

Substituting the value of x and y in given expression, we get;

$⇒ (2a - b)^3 $

Thus, the factorization is

$⇒ (2a - b)(2a - b)(2a - b) $

Question 8 (iii)

Factorize each of the following :
(iii) 27 – 125a³ – 135a + 225a²

Solution :

Given , 27 – 125a³ – 135a + 225a²

This can be re-written as:

$⇒ (3)^3 - (5^3)a^3 $ - $ 135a + 225a^2 $

$⇒ (3)^3 - (5a)^3 $ - $ 135a + 225a^2 $

$⇒ (3)^3 - (5a)^3 $ - $ 45a( 3 - 5a) $

$⇒ (3)^3 - (5a)^3 $ - $ 3 × 3 × 5a ( 3 - 5a) $

Let, x = 3 , y = 5a

Using suitable identity : (x - y)³ = x³ - y³ - 3xy (x - y)

Substituting the value of x and y in given expression, we get;

$⇒ (3 - 5a)^3 $

Thus, the factorization is

$⇒ (3 - 5a)(3 - 5a)(3 - 5a) $

Question 8 (iv)

Factorize each of the following :
(iv) 64a³ – 27b³ - 144a²b + 108ab²

Solution :

Given , 64a³ – 27b³ - 144a²b + 108ab²

This can be re-written as:

$⇒ (4^3)a^3 - (3^3)b^3 $ - $ 144a^2b + 108ab^2 $

$⇒ (4a)^3 - (3b)^3 $ - $ 144a^2b + 108ab^2 $

$⇒ (4a)^3 - (3b)^3 $ - $ 36ab( 4a - 3b) $

$⇒ (4a)^3 - (3b)^3 $ - $ 3 × 4a × 3b( 4a - 3b) $

Let, x = 4a , y = 3b

Using suitable identity : (x - y)³ = x³ - y³ - 3xy (x - y)

Substituting the value of x and y in given expression, we get;

$⇒ (4a - 3b)^3 $

Thus, the factorization is

$⇒ (4a - 3b)(4a - 3b)(4a - 3b) $

Question 8 (v)

Factorize each of the following :
(v) 27p³ – ${1 \over 216} $ - ${9 \over 2} $p² + ${1 \over 4} $p

Solution :

Given , 27p³ – ${1 \over 216} $ - ${9 \over 2} $p² + ${1 \over 4} $p

This can be re-written as:

$⇒ (3^3)p^3 - ({1^3 \over {6^3}}) $ - $ {9 \over 2}p^2 + {1 \over 4}p $

$⇒ (3p)^3 - ({1 \over 6})^3 $ - ${9 \over 2}p^2 + {1 \over 4}p $

$⇒ (3p)^3 - ({1 \over 6})^3 $ - $ 3 × 3p × {1 \over 6}( 3p - {1 \over 6}) $

Let, x = 3p , y = ${1 \over 6} $

Using suitable identity : (x - y)³ = x³ - y³ - 3xy (x - y)

Substituting the value of x and y in given expression, we get;

$⇒ (3p - {1 \over 6})^3 $

Thus, the factorization is

$⇒ (3p - {1 \over 6})(3p - {1 \over 6})(3p - {1 \over 6}) $

Question 9 (i)

Verify:
(i) x³ + y³ = (x + y) (x² – xy + y²)

Solution :

Given , x³ + y³ = (x + y) (x² – xy + y²)

Consider R.H.S.

(x + y) (x² – xy + y²)

$⇒ x × (x^2 - xy + y^2) + y × (x^2 - xy + y^2) $

Expand by multiplying x with the second trinomial, and then y with the second trinomial:

$⇒ x^3 - x^2y + xy^2 + x^2y - xy^2 + y^3 $

Distribute x and y into the parentheses

$⇒ x^3 - x^2y + x^2y + xy^2 - xy^2 + y^3 $

Group and cancel the like terms

$⇒ x^3 + y^3 $

= L. H.S.

Since R. H.S. = x³ + y³ = L. H.S., the identity is verified.

Question 9 (ii)

Verify:
(ii) x³ - y³ = (x - y) (x² + xy + y²)

Solution :

Given , x³ - y³ = (x - y) (x² + xy + y²)

Consider R.H.S.

(x - y) (x² + xy + y²)

$⇒ x × (x^2 + xy + y^2) - y × (x^2 + xy + y^2) $

Expand by multiplying x with the second trinomial, and then -y with the second trinomial:

$⇒ x^3 + x^2y + xy^2 - x^2y - xy^2 - y^3 $

Distribute x and y into the parentheses

$⇒ x^3 + x^2y - x^2y + xy^2 - xy^2 - y^3 $

Group and cancel the like terms

$⇒ x^3 - y^3 $

= L. H.S.

Since R. H.S. = x³ - y³ = L. H.S., the identity is verified.

Question 10 (i)

Factorise each of the following:
(i) 27 y³ + 125z³

Solution :

Given , 27 y³ + 125z³

This can be re-written as:

$⇒ (3^3)y^3 + (5^3)z^3 $

$⇒ (3y)^3 + (5z)^3 $

Let, a = 3y , b = 5z

Using suitable identity : a³ + b³ = (a + b) (a² – ab + b²)

Substituting the value of a and b in given expression, we get;

$⇒ 27y^3 + 125z^3 $

$⇒ (3y + 5z) [(3y)^2 – (3y × 5z) + (5z)^2] $

$⇒ (3y + 5z) ( 9y^2 – 15yz + 25z^2) $

Thus, the factorization is

(3y + 5z) ( 9y² – 15yz + 25z²)

Question 10 (ii)

Factorise each of the following:
(ii) 64m³ - 343n³

Solution :

Given , 64m³ - 343n³

This can be re-written as:

$⇒ (4^3)m^3 - (7^3)n^3 $

$⇒ (4m)^3 - (7n)^3 $

Let, a = 4m , b = 7n

Using suitable identity : a³ - b³ = (a - b) (a² + ab + b²)

Substituting the value of a and b in given expression, we get;

$⇒ 64m^3 - 343n^3 $

$⇒ (4m - 7n) [(4m )^2 + (4m ×7n) + (7n)^2] $

$⇒ (4m - 7n) ( 16m^2 + 28mn + 49n^2) $

Thus, the factorization is

(4m - 7n) ( 16m² + 28mn + 49n²)

Question 11

Factorise :
27 x³ + y³ + z³ -9xyz

Solution :

Given , 27 x³ + y³ + z³ -9xyz

This can be re-written as:

$⇒ (3^3)x^3 + y^3+ z^3 - 9xyz $

$⇒ (3x)^3 + y^3+ z^3 - 3 × 3x × y × z $

Let, x = 3x , y = y and z = z

Using suitable identity :
x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² – xy – yz - zx)

$⇒ 27x^3 + y^3+ z^3 - 9xyz $

Substituting the value of x and y and z in given expression, we get;

$⇒ (3x + y + z) $ $[(3x)^2 + y^2 + z^2 – (3x × y) – (y×z) – (z×3x)] $

$⇒ (3x + y + z) $ $[9x^2 + y^2 + z^2 – 3xy – yz – 3zx] $

Thus, the factorization is

(3x + y + z) [9x² + y² + z² – 3xy – yz – 3zx]

Question 12

Verify that:
$ x^3 + y^3 + z^3 – 3xyz =$ ${1 \over 2}(x + y + z)$ $[(x – y)^2 + (y – z)^2 + (z – x)^2] $

Solution :

Given , $ x^3 + y^3 + z^3 – 3xyz =$ ${1 \over 2}(x + y + z)$ $[(x – y)^2 + (y – z)^2 + (z – x)^2] $

Consider R.H.S.

$ {1 \over 2}(x + y + z) [(x – y)^2 + (y – z)^2 + (z – x)^2] $

Using suitable identity : (x-y)² = x² - 2xy + y²

$⇒ {1 \over 2}(x + y + z) $ $[(x^2 + y^2 - 2xy) $ + $(y^2 + z^2 - 2yz) + (z^2+ x^2 - 2zx)] $

$⇒ {1 \over 2}(x + y + z) $ $ (2x^2 + 2y^2 + 2z^2- 2xy - 2yz - 2zx) $

$⇒ {1 \over 2}(x + y + z) $ $ 2 ×(x^2 + y^2 + z^2- xy - yz - zx) $

$⇒ (x + y + z) $ $ (x^2 + y^2 + z^2- xy - yz - zx) $

We know that

x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² – xy – yz - zx)

Therefore ,

$ x^3 + y^3 + z^3 – 3xyz $

= L.H.S.

Hence Proved

Question 13

If x + y + z = 0, show that
x³ + y³ + z³ = 3xyz

Solution :

We know that,: x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² – xy – yz - zx)

Substituting the value of x + y + z = 0, we get;

$⇒ x^3 + y^3+ z^3 - 3xyz $ $ = (0)(x^2 + y^2+ z^2 – xy – yz - zx) $

$⇒ x^3 + y^3+ z^3 - 3xyz = 0 $

$⇒ x^3 + y^3+ z^3 = 3xyz $

Hence Proved.

Question 14 (i)

Without actually calculating the cubes, find the value of each of the following :
(i) (-12)³ + (7)³ + (5)³

Solution :

Given , (-12)³ + (7)³ + (5)³

Let, x = -12 , y = 7 and z = 5

Now , (x + y + z) = (-12) + (7) + (5) = 0

We know that,: if x + y + z = 0 then,

x³ + y³ + z³ = 3xyz

Substituting the value of x and y and z in given expression, we get;

$⇒ (-12)^3 + 7^3+ 5^3 $ $= 3 × (-12) × 7 × 5 $

$⇒ 3 × (-420) $

$⇒ - 1260 $

Therefore, The value of

(-12)³ + (7)³ + (5)³ = -1260

Question 14 (ii)

Without actually calculating the cubes, find the value of each of the following :
(ii) 28³ + (-15)³ + (-13)³

Solution :

Given , 28³ + (-15)³ + (-13)³

Let, x = 28 , y = -15 and z = -13

Now , (x + y + z) = (28) + (-15) + (-13) = 0

We know that,: if x + y + z = 0 then,

x³ + y³ + z³ = 3xyz

Substituting the value of x and y and z in given expression, we get;

$⇒ 28^3 + (-15)^3+ (-13)^3 $ $ = 3 × 28 × (-15) × (-13) $

$= 84 × 195 $

$⇒ 16380 $

Therefore, The value of

28³ + (-15)³ + (-13)³ = 16380

Question 15 (i)

Given possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
(i) Area: 25a² - 35a + 12

Solution :

Given , Area: 25a² - 35a + 12

We know that,: Area of rectangle = Length × Breadth ,

Hence, First we shall factorise the given expression

Given , 25a² - 35a + 12

For quadratic polynomial, We factorize By Splitting the middle term method:

Find 2 numbers p,q such that :

(i) p × q = product of the co-efficient of a² and the constant term (last term)

(ii) p + q = co-efficient of middle term or a

p × q = 25 × 12 = 300

p + q = -35

By trial and error method,

We get p = -20 and q = -15

Now, splitting the middle term of given polynomial can be written as follows:

$$ = 25a^2 - 20a -15a + 12 $$

( By taking 5a and -3 as common, we get)

$$ = 5a (5a - 4) - 3 (5a - 4) $$

( By taking (5a - 4) as common, we get)

$$ = (5a - 4) (5a - 3) $$

Factors of given polynomial : (5a - 4)(5a - 3)

Therefore,

Length = (5a - 4) and Breadth = (5a - 3)

Or Length = (5a - 3) and Breadth = (5a - 4)

Question 15 (ii)

Given possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
(ii) Area: 35y² + 13y - 12

Solution :

Given , Area: 35y² + 13y - 12

We know that,: Area of rectangle = Length × Breadth ,

Hence, First we shall factorise the given expression

Given , 35y² + 13y - 12

For quadratic polynomial, We factorize By Splitting the middle term method:

Find 2 numbers p,q such that :

(i) p × q = product of the co-efficient of a² and the constant term (last term)

(ii) p + q = co-efficient of middle term or a

p × q = 35 × -12 = -420

p + q = 13

By trial and error method,

We get p = 28 and q = -15

Now, splitting the middle term of given polynomial can be written as follows:

$$ = 35y^2 + 28y - 15y - 12 $$

( By taking 7y and -3 as common, we get)

$$ = 7y (5y + 4) - 3 (5y + 4) $$

( By taking (5y + 4) as common, we get)

$$ = (5y + 4) (7y - 3) $$

Factors of given polynomial : (5y + 4) (7y - 3)

Therefore,

Length = (5y + 4) and Breadth = (7y - 3)

Or Length = (7y - 3) and Breadth = (5y + 4)

Question 16 (i)

What are the possible expressions for the dimensions of the cuboids whose volume are given below:
(i) Volume: 3x² -12x

Solution :

Given , Volume: 3x² -12x

We know that,: Volume of a cubiod = length × breadth × height

Hence, First we shall factorise the given expression

$⇒ 3x^2 -12x $

$= 3x × (x - 4x) $

$= 3 × x × (x - 4x) $

Therefore,

Possible length, breadth and height are 3x , x , (x-4)

Question 16 (ii)

What are the possible expressions for the dimensions of the cuboids whose volume are given below:
(ii) Volume: 12ky² + 8 ky – 20k

Solution :

Given , Volume: 12ky² + 8 ky – 20k

We know that,: Volume of a cubiod = length × breadth × height

Hence, First we shall factorise the given expression

$⇒ 12ky^2 + 8 ky – 20k $

$= 4k × (3y^2 + 2y - 5) $

$= 4k × (3y^2 – 3y + 5y – 5) $

$= 4k × [3y (y – 1) + 5(y – 1)] $

$= 4k × (y – 1)× (3y + 5) $

Therefore,

Possible length, breadth and height are 4k , (y – 1),(3y + 5)