Question Number 7008 by Tawakalitu. last updated on 05/Aug/16
$${Factorize}:\:\:\left({a}\:−\:{b}\right)^{\mathrm{5}} \:+\:\left({b}\:−\:{c}\right)^{\mathrm{5}} \:+\:\left({c}−\:{a}\right)^{\mathrm{5}} \\ $$
Commented by Yozzii last updated on 05/Aug/16
$$\left({a}−{b}\right),\:\left({b}−{c}\right)\:{and}\:\left({c}−{a}\right)\:{are}\:{factors} \\ $$$${since}\:{the}\:{expression}\:{above}\:{reduces} \\ $$$${to}\:{zero}\:{if}\:{a}={b},\:{or}\:{b}={c}\:{or}\:{c}={a}. \\ $$$$\Rightarrow\left({a}−{b}\right)^{\mathrm{5}} +\left({b}−{c}\right)^{\mathrm{5}} +\left({c}−{a}\right)^{\mathrm{5}} \\ $$$$=\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right){f}\left({a},{b},{c}\right) \\ $$$${where}\:{f}\left({a},{b},{c}\right)\:{is}\:{an}\:{expression}\:{in}\:{a},{b}\:{and}\:{c}. \\ $$
Commented by sou1618 last updated on 06/Aug/16
$$\left({a}−{b}\right)^{\mathrm{5}} +\left({b}−{c}\right)^{\mathrm{5}} +\left({c}−{a}\right)^{\mathrm{5}} =\frac{\mathrm{5}}{\mathrm{2}}\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)\left\{\left({a}−{b}\right)^{\mathrm{2}} +\left({b}−{c}\right)^{\mathrm{2}} +\left({c}−{a}\right)^{\mathrm{2}} \right\} \\ $$
Commented by Tawakalitu. last updated on 06/Aug/16
$${Thanks}\:{very}\:{much} \\ $$
Answered by Yozzii last updated on 09/Aug/16
$${Check}\:{for}\:{an}\:{answer}\:{in}\:{comments}. \\ $$