Math, asked by Anonymous, 10 months ago

Without finding the actual cubes, factorise (x-2y)^3+(2y-3z)^3+(3z-x)^3.

experts answer pls.​

Answers

Answered by PraptiMishra05
12

"Factorise (x – 2y)3 + (2y – 3z)3 + (3z – x)3 ", then

here is the solution :-

Consider (x – 2y) + (2y – 3z) + (3z – x) = x – x – 2y + 2y – 3z + 3z = 0

Using if A + B + C = 0 then A3 + B3 + C3 = 3ABC

∴ (x – 2y)3 + (2y + 3z)3 + (3z – x)3 = 3 (x – 2y) (2y – 3z) (3z – x)

I'm not an expert but hope my answer will help you and if it didn't help you so please report it

Thank you :)

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