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

1 Answer
Jan 4, 2018

Please refer to the Explanation.

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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