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Question 38
Without finding the cubes, factorise:
(x2y)3+(2y3z)3+(3zx)3


Solution

We know that,

a3+b3+c33abc=(a+b+c)(a2+b2+c2abbcca)

Also, if a + b + c = 0,
then a3+b3+c3=3abc .....condition(1)

Here, we see that
a+b+c=(x2y)+(2y3z)+(3zx)=0

Using condition (1), we get(x2y)3+(2y3z)3+(3zx)3
=3(x2y)(2y3z)(3zx)

Mathematics

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