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Factorize-a-b-5-b-c-5-c-a-5-




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

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