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If $$x^2 + x + 1 = 0$$, then the value of $$x^{49} + x^{50} + x^{51} + x^{52} + x^{53}$$ is
From $$x^2 + x + 1 = 0$$ we get $$x^3 - 1 = (x - 1)(x^2 + x + 1) = 0$$, so $$x^3 = 1$$. Since $$48$$, $$51$$ are multiples of 3, the powers reduce to $$x + x^2 + 1 + x + x^2$$. Using $$x^2 + x + 1 = 0$$ this equals $$0 + (x + x^2) = -1$$.
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