Question Number 184505 by SEKRET last updated on 07/Jan/23
$$\:\:\:\frac{\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)^{\mathrm{5}} −\boldsymbol{\mathrm{a}}^{\mathrm{5}} −\boldsymbol{\mathrm{b}}^{\mathrm{5}} −\boldsymbol{\mathrm{c}}^{\mathrm{5}} }{\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)^{\mathrm{3}} −\boldsymbol{\mathrm{a}}^{\mathrm{3}} −\boldsymbol{\mathrm{b}}^{\mathrm{3}} −\boldsymbol{\mathrm{c}}^{\mathrm{3}} }\:=\:? \\ $$$$\:\mathrm{is}\:\mathrm{there}\:{an}\:\:{easier}\:\:{way}. \\ $$
Answered by Frix last updated on 07/Jan/23
$$=\frac{\mathrm{5}\left({a}+{b}\right)\left({a}+{c}\right)\left({b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{ab}+{ac}+{bc}\right)}{\mathrm{3}\left({a}+{b}\right)\left({a}+{c}\right)\left({b}+{c}\right)} \\ $$$$\mathrm{I}\:\mathrm{found}\:\mathrm{no}\:\mathrm{easier}\:\mathrm{way} \\ $$
Commented by SEKRET last updated on 08/Jan/23
$$\:\:\boldsymbol{\mathrm{thanks}}\:\:\boldsymbol{\mathrm{sir}} \\ $$