Using factor theorem, show that $a - b, b - c$ and $c - a$ are the factors of $a (b^{2} - c^{2}) + b {(c^{2} - a)}^{2}$ $c (a^{2} - b^{2})$

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Solution

$a (b^{2} - c^{2}) + b (c^{2} - a^{2}) + c (a^{2} - b^{2})$

$\Rightarrow a b^{2} - a c^{2} + b c^{2} - b a^{2} + c a^{2} - c b^{2}$

$\Rightarrow a b^{2} - b a^{2} - a c^{2} + b c^{2} - c b^{2} + c a^{2}$

$\Rightarrow a b^{2} - b a^{2} + b c^{2} - a c^{2} - c b^{2} + c a^{2}$

$\Rightarrow a b (b - a) + c^{2} (b - a) - c (b^{2} - a^{2})$

$\Rightarrow a b (b - a) + c^{2} (b - a) - c (b + a) (b - a)$

$\Rightarrow (b - a) (a b + c^{2} - c (b + a))$

$\Rightarrow (b - a) (a b + c^{2} - c b - a c)$

$\Rightarrow (b - a) (a b - b c - a c + c^{2})$

$\Rightarrow (b - a) (b (a - c) - c (a - c))$

$\Rightarrow (b - a) (a - c) (b - c)$

$\Rightarrow (a - b) (b - c) (c - a)$ Hence proof

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a - b

a (b^{2} - c^{2}) + b (c^{2} - a^{2}) + c (a^{2} - b^{2})

a + b + c = a b c

\frac{a (b^{2} c^{2} - 1)}{b c + 1} + \frac{b (c^{2} a^{2} - 1)}{c a + 1} + \frac{c (a^{2} b^{2} - 1)}{a b + 1} = 2 a b c

\frac{a (b^{2} + c^{2}) + b (c^{2} + a^{2}) + c (a^{2} + b^{2})}{a b c}

(a^{2} - b^{2}) (b^{2} + c^{2}) = (b^{2} - c^{2}) (a^{2} + b^{2})

a (b^{2} + c^{2}) = c (a^{2} + b^{2})

a (b^{2} + c^{2}) + b (c^{2} + a^{2}) + c (a^{2} + b^{2})

λ a b c

λ

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