Question

without finding cubes factorise (x-2y)^3-(2y-3y)^3+(3y-x)^3

Answer 1

Given ( x - 2y)3 + (2y - 3z)3 + ( 3z - x)3
Let a =  ( x - 2y), b =(2y - 3z), c= ( 3z - x)
a + b + c = ( x - 2y)+(2y - 3z)+( 3z - x) = 0
Recall that if (a + b + c) = 0 then a3 + b3 + c3 = 3abc
Thus, ( x - 2y)3 + (2y - 3z)3 + ( 3z - x)3 = 3( x - 2y)(2y - 3z)( 3z - x)

Answer 2

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

= $<b>Using ( a + b + c ) = 0 then a³ + b³ + c³ = 3abc </b>$

$=\: Here \:\: a \: = \: x \:- \:2y\\\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:b \: = \: 2y \: - \: 3y \\\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:c \: = \: 3y \:-\:x$

= a + b + c

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

= x - 2y + 2y - 3y + 3y - x

= 0

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

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

$Please \: mark \: this \: answer \: as \: the \\ BRILLIANIST\\ ANSWER.$