Prove that #a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)#. How can I solve this without expanding everything? Thx

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Answer 1 · 2018-01-04T08:58:22

Please refer to the Explanation.

It is known that, #(a+b)^3=a^3+b^3+3ab(a+b)#.

#:. a^3+b^3=(a+b)^3-3ab(a+b)..............................(star)#.

Setting, #(a+b)=d," we have, "a^3+b^3=d^3-3abd#.

#:. ul(a^3+b^3)+c^3-3abc#,

#=d^3-3abd+c^3-3abc#,

#=ul(d^3+c^3)-ul(3abd-3abc)#,

#=ul((d+c)^3-3dc(d+c))-3ab(d+c)............[because, (star)]#,

#=(d+c)^3-3(d+c)(dc+ab)#,

#=(d+c){(d+c)^2-3(dc+ab)}#,

#=(d+c){d^2+2dc+c^2-3dc-3ab}#,

#=(d+c){d^2+c^2-dc-3ab}#,

#=(a+b+c){(a+b)^2+c^2-(a+b)c-3ab}......[because, d=a+b],#

#=(a+b+c){ul(a^2+2ab+b^2)+c^2-ac-bc-3ab}#.

#=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)#, as desired!

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