If w≠1 is a cube root of unity, x=a+b, y=aw+bw^2 and z=aw^2 +bw, then x^3 +y^3 +z^3 =?


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Question Number 139490 by EnterUsername last updated on 27/Apr/21

$If w \neq 1 is a cube root of unity, x = a + b, y = a w + {b w}^{2}$ $and z = {a w}^{2} + b w, then x^{3} + y^{3} + z^{3} = ?$
	Answered by Rasheed.Sindhi last updated on 27/Apr/21
	$x^3 +y^3 +z^3 =(a+b)^3 +(aw+bw^2 )^3 +(aw^2 +bw)^3 =a^3 +b^3 +3ab(a+b) +a^3 w^3 +b^3 (w^3 )^2 +3abw^3 (aw+bw^2 ) +a^3 (w^3 )^2 +b^3 w^3 +3abw^3 (aw^2 +bw) =a^3 +b^3 +3ab(a+b) +a^3 (1)+b^3 (1)^2 +3ab(1)(aw+bw^2 ) +a^3 (1)^2 +b^3 (1)+3ab(1)(aw^2 +bw) =3a^3 +3b^3 +3ab(a+aw+aw^2 +b+bw+bw^2 ) =3a^3 +3b^3 +3ab{a(1+w+w^2 )+b(1+w+w^2 )} =3a^3 +3b^3 +3ab{a(0)+b(0)} =3(a^3 +b^3 )$
	$x^{3} + y^{3} + z^{3} = {(a + b)}^{3} + {(a w + {b w}^{2})}^{3} + {({a w}^{2} + b w)}^{3}$ $= a^{3} + b^{3} + 3 a b (a + b)$ $+ a^{3} w^{3} + b^{3} {(w^{3})}^{2} + 3 {a b w}^{3} (a w + {b w}^{2})$ $+ a^{3} {(w^{3})}^{2} + b^{3} w^{3} + 3 {a b w}^{3} ({a w}^{2} + b w)$ $= a^{3} + b^{3} + 3 a b (a + b)$ $+ a^{3} (1) + b^{3} {(1)}^{2} + 3 a b (1) (a w + {b w}^{2})$ $+ a^{3} {(1)}^{2} + b^{3} (1) + 3 a b (1) ({a w}^{2} + b w)$ $= 3 a^{3} + 3 b^{3} + 3 a b (a + a w + {a w}^{2} + b + b w + {b w}^{2})$ $= 3 a^{3} + 3 b^{3} + 3 a b {a (1 + w + w^{2}) + b (1 + w + w^{2})}$ $= 3 a^{3} + 3 b^{3} + 3 a b {a (0) + b (0)}$ $= 3 (a^{3} + b^{3})$
	Commented by EnterUsername last updated on 27/Apr/21

	$T h a n k s S i r$
	Answered by Ar Brandon last updated on 27/Apr/21

	$x^{3} + y^{3} + z^{3} - 3 xy z = (x + y + z) (x + y w + {z w}^{2}) (x + y w^{2} + z w)$ $x^{3} + y^{3} + z^{3} - 3 (a + b) (a w + {b w}^{2}) ({a w}^{2} + b w) = 0$ $x^{3} + y^{3} + z^{3} = 3 (a + b) (a^{2} w^{3} + b^{2} w^{3} + a b (w^{2} + w^{4})$ $x^{3} + y^{3} + z^{3} = 3 (a + b) (a^{2} - a b + b^{2}) = 3 (a^{3} + b^{3})$
	Commented by Ar Brandon last updated on 27/Apr/21

	$x + y + z = a + b + (w + w^{2}) a + (w + w^{2}) b$ $= a + b - a - b = 0 (w^{2} + w = - 1), w^{4} = w$
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