# Show that a^2b^2(a-b)+b^2c^2(b-c)+c^2a^2(c-a) us exactly divisible by a-b,b-c and c-a?

Jun 11, 2018

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

$= {a}^{2} {b}^{2} \left(a - b\right) + {b}^{3} {c}^{2} - {b}^{2} {c}^{3} + {c}^{3} {a}^{2} - {c}^{2} {a}^{3}$

$= {a}^{2} {b}^{2} \left(a - b\right) - {c}^{2} \left({a}^{3} - {b}^{3}\right) + {c}^{3} \left({a}^{2} - {b}^{2}\right)$

$= \left(a - b\right) \left({a}^{2} {b}^{2} - {c}^{2} \left({a}^{2} + a b + {b}^{2}\right) + {c}^{3} \left(a + b\right)\right)$

$= \left(a - b\right) \left({a}^{2} {b}^{2} - {c}^{2} {a}^{2} - {c}^{2} a b - {c}^{2} {b}^{2} + {c}^{3} a + {c}^{3} b\right)$

$= \left(a - b\right) \left({a}^{2} \left({b}^{2} - {c}^{2}\right) - {c}^{2} a \left(b - c\right) - {c}^{2} b \left(b - c\right)\right)$

$= \left(a - b\right) \left(b - c\right) \left({a}^{2} b + {a}^{2} c - {c}^{2} a - {c}^{2} b\right)$

$= \left(a - b\right) \left(b - c\right) \left(b \left({a}^{2} - {c}^{2}\right) + a c \left(a - c\right)\right)$

$= \left(a - b\right) \left(b - c\right) \left(a - c\right) \left(a b + b c + c a\right)$

This factorisation proves that the given expression is exactly divisible by $\left(a - b\right) , \left(b - c\right) \mathmr{and} \left(c - a\right)$