Using factor theorem, show that $a - b$ is a factor of $a (b^{2} - c^{2}) + b (c^{2} - a^{2}) + c (a^{2} - b^{2})$

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Solution

We know that the factor theorem states that if the polynomial $p (x)$ is divided by $(c x - d)$ and the remainder, given by $p (\frac{d}{c})$ , is equal to zero, then $(c x - d)$ is a factor of $p (x)$ .

Consider the given expression $a (b^{2} - c^{2}) + b (c^{2} - a^{2}) + c (a^{2} - b^{2})$ and solving it as follows:

$a (b^{2} - c^{2}) + b (c^{2} - a^{2}) + c (a^{2} - b^{2}) = a b^{2} - a c^{2} + b c^{2} - b a^{2} + c (a - b) (a + b) (∵ (x + y) (x - y) = x^{2} - y^{2}) = a b^{2} - b a^{2} - a c^{2} + b c^{2} + c (a - b) (a + b) = a b (b - a) - (a - b) c^{2} + c (a - b) (a + b) = - a b (a - b) - (a - b) c^{2} + c (a - b) (a + b) = (a - b) (- a b - c^{2} + c (a + b)) = (a - b) (c (a + b) - a b - c^{2})$

Hence, by factor theorem we have proved that $(a - b)$ is a factor of $a (b^{2} - c^{2}) + b (c^{2} - a^{2}) + c (a^{2} - b^{2})$ .

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