Without evaluating the cubes, factorize
(I) (x – y)3 + (y – z)3 + (z – x)3
(II) (x – 2y)3 + (2y – 3z)3 + (3z – x)3
(I) (x – y)3 + (y – z)3 + (z – x)3
As the expression is of the form of sum of three cubes, for factorizing, we will use the identity
a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc –ca)
Taking the 3abc term to the right hand side, we have
a3 + b3 + c3 = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) + 3abc
Here, a = (x – y), b = (y – z) & c = (z – x)
⇒ a3 + b3 + c3 = (x – y)3 + (y – z)3 + (z – x)3
Observe that a + b + c = (x – y) + (y – z) + (z – x) = 0
⇒ a3 + b3 + c3 = 0 + 3abc = 3abc
∴ (x – y)3 + (y – z)3 + (z – x)3 = 3(x – y)(y – z)(z – x)
(II)(x – 2y)3 + (2y – 3z)3 + (3z – x)3
As the expression is again of the form of sum of three cubes, for
factorizing, we will use the same identity as above
a3 + b3 + c3 = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) + 3abc
Here, a = (x – 2y), b = (2y – 3z) & c = (3z – x)
⇒ a3 + b3 + c3 = (x – 2y)3 + (2y – 3z)3 + (3z – x)3
Observe that a + b + c = (x – 2y) + (2y – 3z) + (3z – x) = 0
⇒ a3 + b3 + c3 = 0 + 3abc = 3abc
∴ (x – 2y)3 + (2y – 3z)3 + (3z – x)3 = 3(x – 2y)(2y – 3z)(3z – x)
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