Question

If $\alpha, \beta, \gamma$ are roots of equation $x^{3}-2 x^{2}-1=0$ and $T _{ n }=\alpha^{ n }+\beta^{ n }+\gamma^{ n }$, then value of $\frac{ T _{11}- T _{8}}{ T _{10}}$ is equal to

• A. 1
• B. 2
• C. -1
• D. 3

Answer 1

Answer: (B)

$x^{3}-2 x^{2}-1=0$ or $\alpha^{3}-1=2 \alpha^{2}$
$\therefore \frac{T_{11}-T_{8}}{T_{10}}=\frac{\alpha^{11}+\beta^{11}+\gamma^{11}-\left[\alpha^{8}+\beta^{8}+\gamma^{8}\right]}{\alpha^{10}+\beta^{10}+\gamma^{10}}$
$=\frac{\alpha^{8}\left(\alpha^{3}-1\right)+\beta^{8}\left(\beta^{3}-1\right)+\gamma^{8}\left(\gamma^{3}-1\right)}{\alpha^{10}+\beta^{10}+\gamma^{10}}$
$=\frac{2\left(\alpha^{10}+\beta^{10}+\gamma^{10}\right)}{\alpha^{10}+\beta^{10}+\gamma^{10}}=2$