Q. 155.0( 4 Votes )

# Factorize (x – 2y)^{3} + (2y – 3z)^{3} + (3z – x)^{3} without finding the cubes explicitly.

Answer :

As the expression is of the form of sum of three cubes, for factorizing, we will use the identity

a^{3} + b^{3} + c^{3} – 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} – ab – bc – ca)

Taking the 3abc term to the right hand side, we have

a^{3} + b^{3} + c^{3} = (a + b + c)(a^{2} + b^{2} + c^{2} – ab – bc – ca) + 3abc

Here, a = (x – 2y), b = (2y – 3z) & c = (3z – x)

⇒ a^{3} + b^{3} + c^{3} = (x – 2y)^{3} + (2y – 3z)^{3} + (3z – x)^{3}

Observe that a + b + c = (x – 2y) + (2y – 3z) + (3z – x) = 0

⇒ a^{3} + b^{3} + c^{3} = 0 + 3abc = 3abc

∴ (x – 2y)^{3} + (2y – 3z)^{3} + (3z – x)^{3} = 3(x – 2y)(2y – 3z)(3z – x)

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