The quadratic equation whose roots are ${{\left( \dfrac{\alpha }{\gamma } \right)}^{3}}$ and ${{\left( \dfrac{\beta }{\gamma } \right)}^{3}}$, where $\alpha ,\beta ,\gamma $ are the roots of the equation ${{x}^{3}}-8=0$, is
(A) ${{x}^{2}}+x+1=0$
(B) ${{x}^{2}}+2x+4=0$
(C) ${{x}^{2}}-2x+4=0$
(D) ${{x}^{2}}-2x+1=0$

Hint: We solve this equation by substituting the roots $\alpha ,\beta ,\gamma $ in the equation ${{x}^{3}}-8=0$. Then we use the values obtained to find the values of ${{\left( \dfrac{\alpha }{\gamma } \right)}^{3}}$ and ${{\left( \dfrac{\beta }{\gamma } \right)}^{3}}$. As they are roots of the quadratic equation that we have to find, we can use the obtained values to find the equation of the quadratic equation.

Complete step-by-step answer:
We are given that $\alpha ,\beta ,\gamma $ are the roots of the equation ${{x}^{3}}-8=0$. As they are the roots of the equation, they need to satisfy the equation ${{x}^{3}}-8=0$. So, by substituting them in the equation we get,
$\begin{align}
  & \Rightarrow {{\alpha }^{3}}=8 \\
 & \Rightarrow {{\beta }^{3}}=8 \\
 & \Rightarrow {{\gamma }^{3}}=8 \\
\end{align}$
As we need to find the equation with roots ${{\left( \dfrac{\alpha }{\gamma } \right)}^{3}}$ and ${{\left( \dfrac{\beta }{\gamma } \right)}^{3}}$, let us substitute the values obtained in the them. Then we get,
$\begin{align}
  & \Rightarrow {{\left( \dfrac{\alpha }{\gamma } \right)}^{3}}=\dfrac{{{\alpha }^{3}}}{{{\gamma }^{3}}}=\dfrac{8}{8}=1 \\
 & \Rightarrow {{\left( \dfrac{\beta }{\gamma } \right)}^{3}}=\dfrac{{{\beta }^{3}}}{{{\gamma }^{3}}}=\dfrac{8}{8}=1 \\
\end{align}$
So, we get that both the values are equal to 1.
So, both the roots of the equation are 1. As they are the roots of the quadratic equation, we can write the quadratic equation as,
$\begin{align}
  & \Rightarrow \left( x-1 \right)\left( x-1 \right)=0 \\
 & \Rightarrow {{x}^{2}}-2x+1=0 \\
\end{align}$
Hence, the quadratic equation whose roots are ${{\left( \dfrac{\alpha }{\gamma } \right)}^{3}}$ and ${{\left( \dfrac{\beta }{\gamma } \right)}^{3}}$ is ${{x}^{2}}-2x+1=0$.

So, the correct answer is “Option D”.

Note: One can also solve this problem by taking the sum and product of the roots of the equation ${{x}^{3}}-8=0$.
$\begin{align}
  & \Rightarrow \alpha +\beta +\gamma =0 \\
 & \Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =0 \\
 & \Rightarrow \alpha \beta \gamma =8 \\
\end{align}$
Then we need to find the values of ${{\left( \dfrac{\alpha }{\gamma } \right)}^{3}}+{{\left( \dfrac{\beta }{\gamma } \right)}^{3}}$ and ${{\left( \dfrac{\alpha }{\gamma } \right)}^{3}}\times {{\left( \dfrac{\beta }{\gamma } \right)}^{3}}$. Then we substitute the obtained values as the sum of the roots and product of the roots of the quadratic equation and can find the answer. But it is a long process and takes too much time for solving the question.