Question

If $a$ is a root of the equation $x^2-3x+1=0$ , then the value of $\frac{a^3}{a^6+1}$ is

• A. $\frac{1}{14}$
• B. $\frac{1}{15}$
• C. $\frac{1}{16}$
• D. $\frac{1}{17}$
• E. $\frac{1}{18}$

Answer 1

Answer: (E)

Given, $a$ is a root of the equation
$x^2-3x+1=0$
$\therefore a^2-3a+1=0$
$\Rightarrow a^2+1=3a$
$\Rightarrow a+\frac{1}{a}=3\dots (i)$
Now $\frac{a^3}{a^6+1}=\frac{1}{a^3+\frac{1}{a^3}}$
$=\frac{1}{\left(a+\frac{1}{a}\right)\left(a^2+\frac{1}{a^2}-1\right)}$
$=\frac{1}{\left(a+\frac{1}{a}\right)\left\{{\left(a+\frac{1}{a}\right)}^2-3\right\}}$
$=\frac{1}{3\{9-3\}}=\frac{1}{18}$ [From Eq. (i)]