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
$x^{2}-3 x+1=0$
$\therefore \, a^{2}-3 a+1=0$
$\Rightarrow \, a^{2}+1=3 a$
$\Rightarrow \, a+\frac{1}{a}=3\,\,\,\,\,\,\dots(i)$
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)]