The sum of the complex roots of the equation | vedclass

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Question

(A) Given equation: $(x-1)^3+64=0$ $\Rightarrow (x-1)^3 = -64$ $\Rightarrow (x-1)^3 = (-4)^3$ Let $y = x-1$,then $y^3 = (-4)^3$. The roots are $y = -4

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) Given equation: $(x-1)^3+64=0$ $\Rightarrow (x-1)^3 = -64$ $\Rightarrow (x-1)^3 = (-4)^3$ Let $y = x-1$,then $y^3 = (-4)^3$. The roots are $y = -4, -4\omega, -4\omega^2$,where $\omega$ is the complex cube root of unity. Thus,$x-1 = -4, -4\omega, -4\omega^2$. The roots are $x_1 = -3$,$x_2 = 1-4\omega$,and $x_3 = 1-4\omega^2$. The complex roots are $x_2 = 1-4\omega$ and $x_3 = 1-4\omega^2$. Sum of complex roots $= (1-4\omega) + (1-4\omega^2) = 2 - 4(\omega + \omega^2)$. Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$. Sum $= 2 - 4(-1) = 2 + 4 = 6$.

The sum of the complex roots of the equation $(x-1)^3+64=0$ is

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