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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} + {(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