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Question Number 59060 by Tawa1 last updated on 04/May/19

	Answered by tanmay last updated on 04/May/19
	$a^3 +b^3 +c^3 −3abc (a+b+c)(a^2 +b^2 +c^2 −ab−bc−ca) (a+b+c)(a^2 +b^2 +c^2 +abw+abw^2 +bcw+bcw^2 +caw+caw^2 ) we know w^3 =1 and 1+w+w^2 =0 (a+b+c)(a^2 +abw+acw^2 +b^2 w^3 +abw^2 +bcw^4 +c^2 w^3 +bcw^2 +acw) (a+b+c){a(a+bw+cw^2 )+bw^2 (a+bw+cw^2 )+cw(a+bw+cw^2 )} (a+b+c)(a+bw+cw^2 )(a+bw^2 +cw)$
	$a^{3} + b^{3} + c^{3} - 3 a b c$ $(a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)$ $(a + b + c) (a^{2} + b^{2} + c^{2} + a b w + {a b w}^{2} + b c w + {b c w}^{2} + c a w + {c a w}^{2})$ $w e k n o w w^{3} = 1 a n d 1 + w + w^{2} = 0$ $(a + b + c) (a^{2} + a b w + {a c w}^{2} + b^{2} w^{3} + {a b w}^{2} + {b c w}^{4} + c^{2} w^{3} + {b c w}^{2} + a c w)$ $(a + b + c) {a (a + b w + {c w}^{2}) + {b w}^{2} (a + b w + {c w}^{2}) + c w (a + b w + {c w}^{2})}$ $(a + b + c) (a + b w + {c w}^{2}) (a + {b w}^{2} + c w)$
	Commented by Tawa1 last updated on 04/May/19

	$God bless you sir$
	Answered by math1967 last updated on 04/May/19
	$(a+b+c)(a^2 +abω^2 +caω+abω+b^2 ω^3 +bcω^2 +caω^2 +bcω^3 ω+ω^3 c^2 ) =(a+b+c){a^2 +b^2 +c^2 +ab(ω^2 +ω) +bc(ω^2 +ω)+ca(ω+ω^2 )} =(a+b+c)(a^2 +b^2 +c^2 −ab−bc−ca)★ =a^3 +b^3 +c^3 −3abc ★ω+ω^2 +1=0∴ω+ω^2 =−1 ω^3 =1$
	$(a + b + c) (a^{2} + a b ω^{2} + c a ω + a b ω + b^{2} ω^{3}$ $+ b c ω^{2} + c a ω^{2} + b c ω^{3} ω + ω^{3} c^{2})$ $= (a + b + c) {a^{2} + b^{2} + c^{2} + a b (ω^{2} + ω)$ $+ b c (ω^{2} + ω) + c a (ω + ω^{2})}$ $= (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) ★$ $= a^{3} + b^{3} + c^{3} - 3 a b c$ $★ ω + ω^{2} + 1 = 0 ∴ ω + ω^{2} = - 1$ $ω^{3} = 1$
	Commented by Tawa1 last updated on 04/May/19

	$God bless you sir$
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