Question

(a) x4- (x-y)4 factorise​

Answer 1

Answer:

y(2x – y) (2x2 – 2xy + y2)

Step-by-step explanation:

x4 – (x – y)4

= (x2)2 – [(x – y)2]2

= [x2 – (x – y)2] [x2 + (x – y)2]

= [x + (x – y] [x – (x – y)] [x2 + x2 – 2xy + y2]

= (x + x – y) (x – x + y)[2x2 – 2xy + y2]

= (2x – y) y(2x2 – 2xy + y2)

= y(2x – y) (2x2 – 2xy + y2)

Answer 2

Given:

$x^4-(x-y)^4$

To find:

Factors of $x^4-(x-y)^4$

Solution:

The factors of $x^4-(x-y)^4 \hspace{0.1cm} \text{are}\hspace{0.1cm} (2 x-y) (y)(2 x^2-2 x y+y^2).$

We can solve the above mathematical problem using the following approach.

$x^4-(x-y)^4$ can be written as:

$=(x^2)^2-[(x-y)^2]^2$

This can be further solved into-

$=[x^2-(x-y)^2][x^2+(x-y)^2] \\\\ =[x+(x-y)][x-(x-y)][x^2+x^2-2 x y+y^2] (\because a^2 - b^2 = (a+b)(a-b))\\\\=(x+x-y)(x-x+y)[2 x^2-2 x y+y^2] \\\\ =(2 x-y) (y)(2 x^2-2 x y+y^2)$

Therefore, the factors of $x^4-(x-y)^4 \hspace{0.1cm} \text{are}\hspace{0.1cm} (2 x-y) (y)(2 x^2-2 x y+y^2).$