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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

Factorise $9x^{2} + 42x + 49$.

Expand each of the following,using suitable identities: $(x + 2y + 4z)^2$

Evaluate $(153)^{2}$.

Expand $(3x - 2)(3x - 6)$.

Factorise $: 8 x^{3}+y^{3}+27 z^{3}-18 x y z$

Vedclass Test Series

Exam Paper Generator

Online Exam Module