Question

factorise (x-y)^3 + (y-z)^3 + (z-x)^3 solution​

Answer 1

/* We know that */

$If \: a + b + c = 0 \:then$

$a^{3} + b^{3}+c^{3} = 3abc$

$Here, a = x - y , b = y - z \:and \: c = z - x$

$a + b + c = x - y + y - z + z - x$

$= 0$

$Now, a^{3} + b^{3} + c^{3}$

$= (x-y)^3 + (y-z)^3 + (z-x)^3$

$= 3(x-y) (y-z) (z-x)$

Therefore.,

$\red{ Factors \: of \: (x-y)^3 + (y-z)^3 + (z-x)^3}$

$\green {= 3(x-y) (y-z) (z-x)}$

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