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If $\alpha ,\beta \text{ and }\gamma $ are the roots of the equation ${{x}^{3}}+px+q=0$ then the value of determinant \[\left| \begin{matrix}
   \alpha & \beta & \gamma \\
   \beta & \gamma & \alpha \\
   \gamma & \alpha & \beta \\
\end{matrix} \right|\] is
\[\begin{align}
  & A.p \\
 & B.q \\
 & C.{{p}^{2}}-2q \\
 & D.0 \\
\end{align}\]

Answer Verified Verified
Hint: To solve this question, we will use three basic mathematical value which are given as below:
Sum of roots of a cubic (3 degree) equation is given by $\alpha +\beta +\gamma $ where equation is of the type \[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\]
We also know the formula, sum of roots \[\Rightarrow \alpha +\beta +\gamma =\dfrac{-b}{a}=\dfrac{-\text{coefficient of }{{\text{x}}^{\text{2}}}}{\text{coefficient of }{{\text{x}}^{3}}}\]
Using this, we will get the value of $\alpha +\beta +\gamma $ which can then be substituted in the determinant after applying row transformations.

Complete step-by-step answer:
We are given the equation as ${{x}^{3}}+px+q=0$.
Now, we know that sum of roots of a cubic (3 degree) equation can be written as $\alpha +\beta +\gamma $ where equation is of the type \[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\]. We also have a formula for finding the sum of roots. It is given as below,
\[\Rightarrow \alpha +\beta +\gamma =\dfrac{-b}{a}=\dfrac{-\text{coefficient of }{{\text{x}}^{\text{2}}}}{\text{coefficient of }{{\text{x}}^{3}}}\]
Here, in this given equation \[{{x}^{3}}+px+q=0\] we do not have any term of ${{x}^{2}}$ so, the coefficient of ${{x}^{2}}=0$
Hence, we get the sum of roots of equation \[{{x}^{3}}+px+q=0\] as
\[\begin{align}
  & \alpha +\beta +\gamma =0\text{ as coefficient of }{{\text{x}}^{\text{2}}}=0 \\
 & \alpha +\beta +\gamma =0\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)} \\
\end{align}\]
Let us consider the determinant \[\left| \begin{matrix}
   \alpha & \beta & \gamma \\
   \beta & \gamma & \alpha \\
   \gamma & \alpha & \beta \\
\end{matrix} \right|\]
Determinant property says that, sum of elements of row or column changes the value of determinant. Hence, adding all rows element of above determinant we get:
\[\left| \begin{matrix}
   \alpha & \beta & \gamma \\
   \beta & \gamma & \alpha \\
   \gamma & \alpha & \beta \\
\end{matrix} \right|=\left| \begin{matrix}
   \alpha +\beta +\gamma & \beta & \gamma \\
   \alpha +\beta +\gamma & \gamma & \alpha \\
   \alpha +\beta +\gamma & \alpha & \beta \\
\end{matrix} \right|\]
Using equation (i) we get:
\[\left| \begin{matrix}
   \alpha & \beta & \gamma \\
   \beta & \gamma & \alpha \\
   \gamma & \alpha & \beta \\
\end{matrix} \right|=\left| \begin{matrix}
   0 & \beta & \gamma \\
   0 & \gamma & \alpha \\
   0 & \alpha & \beta \\
\end{matrix} \right|\]
Determinant property says that, if any one row or column is zero then the value of determinant is zero.
\[\left| \begin{matrix}
   0 & \beta & \gamma \\
   0 & \gamma & \alpha \\
   0 & \alpha & \beta \\
\end{matrix} \right|=0\]
Therefore, the value of \[\left| \begin{matrix}
   \alpha & \beta & \gamma \\
   \beta & \gamma & \alpha \\
   \gamma & \alpha & \beta \\
\end{matrix} \right|=0\]
Therefore, the answer is 0,

So, the correct answer is “Option D”.

Note: Another way to solve this question can be, opening the determinant \[\left| \begin{matrix}
   0 & \beta & \gamma \\
   0 & \gamma & \alpha \\
   0 & \alpha & \beta \\
\end{matrix} \right|\] to get the required answer. Consider opening from first column, we get:
\[\begin{align}
  & \left| \begin{matrix}
   0 & \beta & \gamma \\
   0 & \gamma & \alpha \\
   0 & \alpha & \beta \\
\end{matrix} \right|=0\left| \begin{matrix}
   \gamma & \alpha \\
   \alpha & \beta \\
\end{matrix} \right|-0\left| \begin{matrix}
   \beta & \gamma \\
   \alpha & \beta \\
\end{matrix} \right|+0\left| \begin{matrix}
   \beta & \gamma \\
   \gamma & \alpha \\
\end{matrix} \right| \\
 & \Rightarrow \left| \begin{matrix}
   0 & \beta & \gamma \\
   0 & \gamma & \alpha \\
   0 & \alpha & \beta \\
\end{matrix} \right|=0-0-0 \\
 & \Rightarrow \left| \begin{matrix}
   0 & \beta & \gamma \\
   0 & \gamma & \alpha \\
   0 & \alpha & \beta \\
\end{matrix} \right|=0 \\
\end{align}\]
Hence, the value of determinants: \[\left| \begin{matrix}
   \alpha & \beta & \gamma \\
   \beta & \gamma & \alpha \\
   \gamma & \alpha & \beta \\
\end{matrix} \right|=0\]