Q51 / 160 · TS EAMCET 2023 (14 May Shift 2) · Mathematics

If $\alpha, \beta$ are the roots of the equation $x^2+x+1=0$, then $(\alpha+\beta)^2+\left(\alpha^2+\beta^2\right)^2+\left(\alpha^3+\beta^3\right)^2+\ldots+\left(\alpha^{12}+\beta^{12}\right)^2=$

  • A
    48
  • B
    12
  • Correct answer
    C
    24
  • D
    36

Solution

Since, roots of equation $x^2+x+1=0$ are complex cube roots of unity $\begin{aligned} & \therefore(\alpha+\beta)^2+\left(\alpha^2+\beta^2\right)^2+\ldots \ldots . .+\left(\alpha^{12}+\beta^{12}\right)^2 \\ & =\left(\omega+\omega^2\right)^2+\left(\omega^2+\omega^4\right)^2+\ldots \ldots . .+\left(\omega^{12}+\omega^{24}\right)^2 \\ & \because 1+\omega+\omega^2=0 \text { and } \omega^3=1 \\ & \Rightarrow \omega^n+\left(\omega^2\right)^n=2, n \text { is multiple cf } 3 \\ & \omega^n+\left(\omega^2\right)^n=-1, n \text { is not multiple of } 3 \\ & \therefore\left[(-1)^2+(-1)^2+(2)^2\right]+\ldots \ldots \ldots \ldots . .4 \text { times } \\ & =6 \times 4=24\end{aligned}$

Hence, the correct answer is option: 'C'.

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