Prove that $a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)$ $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ . 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, ${(a + b)}^{3} = a^{3} + b^{3} + 3 a b (a + b)$ $(a+b)^3=a^3+b^3+3ab(a+b)$ .

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

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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Prove that a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)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