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a-b-c-5-a-5-b-5-c-5-a-b-c-3-a-3-b-3-c-3-is-there-an-easier-way-




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

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