Polynomials — Maths STD 9 — Question

 $\Big(\text{x}-\frac{1}{\text{x}}\Big)^3=\text{x}^3-\frac{1}{\text{x}^3}-3\not\text{x}\frac{1}{\not\text{x}}\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\text{x}^3-\frac{1}{\text{x}^3}=\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)-\text{x}^3-\frac{1}{\text{x}^3}=0$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)-14=0$
Let $\Rightarrow\text{x}-\frac{1}{\text{x}}=\text{t}$
$\Rightarrow t^3+3 t-14=0$
$\Rightarrow t^3-2 t^2+2 t^2-4 t+7 t-14=0$
$\Rightarrow t(t-2)+2 t(t-2)+7(t-2)=0$
$\Rightarrow(t-2)(t+2 t+7)=0$
$\Rightarrow t^2+2 t+7=0 \text { has no real roots }$
So, $t = 2$ is a solution
$\Rightarrow\text{x}-\frac{1}{\text{x}}=2$
Hence, correct option is $(d).$