Factorise (x-2 y)^{3}+(2 y-3 z)^{3}+(3 z-x)^{ | vedclass

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Question

(A) We use the algebraic identity: If $a+b+c=0$,then $a^3+b^3+c^3=3abc$.Let $a = (x-2y)$,$b = (2y-3z)$,and $c = (3z-x)$.Now,calculate the sum $a+b+c$:

Vedclass · Accepted Answer

$3(x-2 y)(2 y-3 z)(3 z-x)$

Vedclass · Answer

(A) We use the algebraic identity: If $a+b+c=0$,then $a^3+b^3+c^3=3abc$.Let $a = (x-2y)$,$b = (2y-3z)$,and $c = (3z-x)$.Now,calculate the sum $a+b+c$:$a+b+c = (x-2y) + (2y-3z) + (3z-x)$$a+b+c = x - x - 2y + 2y - 3z + 3z = 0$.Since the sum is $0$,the expression $(x-2y)^3 + (2y-3z)^3 + (3z-x)^3$ simplifies to $3abc$.Therefore,the factorised form is $3(x-2y)(2y-3z)(3z-x)$.

Factorise $(x-2 y)^{3}+(2 y-3 z)^{3}+(3 z-x)^{3}$.

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