NCERT Class 9 Mathematics

48 MATHEMATICS

Example 24 : Factorise 8x^3 + 27y^3 + 36x^2 y + 54xy^2

Solution : The given expression can be written as

(2x)^3 + (3y)^3 + 3(4x^2 )(3y) + 3(2x)(9y^2 )

=(2x)^3 + (3y)^3 + 3(2x)^2 (3y) + 3(2x)(3y)^2

=(2x + 3y)^3 (Using Identity VI)

=(2x + 3y)(2x + 3y)(2x + 3y)

Now consider (x + y + z)(x^2 + y^2 + z^2 – xy – yz – zx)

On expanding, we get the product as

x(x^2 + y^2 + z^2 – xy – yz – zx) + y(x^2 + y^2 + z^2 – xy – yz – zx)

• 
  

  z(x^2 + y^2 + z^2 – xy – yz – zx) = x^3 + xy^2 + xz^2 – x^2 y – xyz – zx^2 + x^2 y

  
  


• 
  

  y^3 + yz^2 – xy^2 – y^2 z – xyz + x^2 z + y^2 z + z^3 – xyz – yz^2 – xz^2

  
  

= x^3 + y^3 + z^3 – 3xyz (On simplification)

So, we obtain the following identity:

Identity VIII : x^3 + y^3 + z^3 – 3xyz = (x + y + z)(x^2 + y^2 + z^2 – xy – yz – zx)

Example 25 : Factorise : 8x^3 + y^3 + 27z^3 – 18xyz

Solution : Here, we have

8 x^3 + y^3 + 27z^3 – 18xyz

=(2x)^3 + y^3 + (3z)^3 – 3(2x)(y)(3z)

=(2x + y + 3z)[(2x)^2 + y^2 + (3z)^2 – (2x)(y) – (y)(3z) – (2x)(3z)]

=(2x + y + 3z) (4x^2 + y^2 + 9z^2 – 2xy – 3yz – 6xz)

EXERCISE 2.5

1. Use suitable identities to find the following products:
   (i) (x + 4) (x + 10) (ii) (x + 8) (x – 10) (iii) (3x + 4) (3x – 5)

(iv) (y^2 +

3

2 ) (y

(^2) –^3
2 ) (v) (3 – 2x) (3 + 2x)

2. Evaluate the following products without multiplying directly:
   (i) 103 × 107 (ii) 95 × 96 (iii)104 × 96


3. Factorise the following using appropriate identities:

(i) 9 x^2 + 6xy + y^2 (ii) 4y^2 – 4y + 1 (iii)x^2 –

2

100

y