Factorise $(2x-3y)^3+(3y-4z)^3+(4z-2x)^3$

Given: $(2x-3y)^3+(3y-4z)^3+(4z-2x)^3$.

To find: We have to factorize the given expression.

Solution:

$(2x-3y)^3+(3y-4z)^3+(4z-2x)^3$

Using the formula $a^3 + b^3 + c^3 = 3abc$ if $a+b+c = 0$

let $a =2x-3y; b = 3y - 4z; c = 4z - 2x$

$a + b + c = 2x -3y + 3y - 4z + 4z - 2x = 0$

$a^3 + b^3 + c^3 = 3abc$

So $(2x-3y)^3+(3y-4z)^3+(4z-2x)^3 = 3(2x-3y) (3y-4z)(4z-2x)$

So $(2x-3y)^3+(3y-4z)^3+(4z-2x)^3$ has been factorized as shown above.

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Updated on: 10-Oct-2022

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