Question

If $\alpha, \beta, \gamma$ are roots of $x^{3}-5 x+4=0$, then $\left(\alpha^{3}+\beta^{3}+\gamma^{3}\right)^{2}$ is equal to

• A. 12
• B. 13
• C. 169
• D. 144

Answer 1

Answer: (D)

Given, the roots of
$x^{3}-5 x+4 =0 $ are $\alpha, \beta \text { and } \gamma $
$ \therefore \alpha+\beta+\gamma =0$
$ \alpha \beta+\beta \gamma+\gamma \alpha =5 $ and $\alpha \beta \gamma=-4 $
Since, $ \alpha+\beta+\gamma =0 $
$ \therefore \alpha^{3}+\beta^{3}+\gamma^{3} =3 \alpha \beta \gamma$
$=3 \times(-4)=-12 $
Hence, $\left(\alpha^{3}+\beta^{3}+\gamma^{3}\right)^{2} =(-12)^{2}=144 $