Question

If $\alpha$ is the root of the equation $x^2-x+2=0$ then the value of $\frac{6\left(-\alpha^3+2 \alpha^2-\alpha\right)}{\alpha^5-3 \alpha^4+3 \alpha^3-\alpha^2}$ is equal to

• A. 3
• B. 6
• C. 9
• D. 12

Answer 1

Answer: (A)

$\alpha^2-\alpha+2=0$
$\text { So, } \frac{6\left(-\alpha^3+2 \alpha^2-\alpha\right)}{\alpha^5-3 \alpha^4+3 \alpha^3-\alpha^2}=\frac{-6 \alpha\left(\alpha^2-2 \alpha+1\right)}{\alpha^2\left(\alpha^3-3 \alpha^2+3 \alpha-1\right)} $
$=\frac{-6 \alpha(\alpha-1)^2}{\alpha^2(\alpha-1)^3}=\frac{-6}{\alpha(\alpha-1)}=\frac{-6}{\alpha^2-\alpha}=\frac{-6}{-2}=3$ From equation (1)