Question

factorise each of the following using identities 8x^3-(2x-y)^3

Answer 1

Given, 8x^3 - (2x - y) ^3
= (2x) ^3 - (2x - y) ^3
Using the formula:
a^3 - b^3= (a - b)(a^2 + ab + b^2) -------- eq.(1)
Putting the value of a = 2x, b = (2x – y) in the above formula; we get
= (2x – (2x+y)) {(2x) ^2 - 2x*(2x - y) + (2x - y) ^2}
= y {(4x^2 - 4x^2 + 2xy + 4x^2 - 4xy + y^2)}
= y (4x^2 - 2xy + y^2)
Therefore, factorized form of the given problem is y (4x^2 - 2xy + y^2).

Answer 2

Given an algebraic equation :

.
8x^3 - ( 2x - y )^3

.
=( 2x )^3 - ( 2x - y )^3

.
use identity:
----------------

.
a^3 - b^3
------------

.
= ( a - b ) ( a^2 + ab + b^2 )

.
here we have ,

.
a = 2x , b = (2x - y )

so now put the value of a and b in above identity :

.
(2x)^3 - ( 2x - y )^3

.
= [2x - (2x - y)] [( 2x)^2 + 2x ( 2x - y ) + (2x- y)^2]

.
= (2x - 2x + y)( 4x^2 + 4x^2 - 2xy + 4x^2 + y^2 - 4xy )

.
= y ( 12x^2 + y^2 - 6xy )

.
= y [ 12x^2 + y ( y - 6x ) ]

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