Question

If $\alpha ,\beta ,\gamma$ are the roots of the equation ${x^3} + px + q = 0$ then the value of the determinant \left| {\begin{array}{*{20}{c}} \alpha &\beta &\gamma \\\ \beta &\gamma &\alpha \\\ \gamma &\alpha &\beta \end{array}} \right| is
A. $q$
B. 0
C. $p$
D. ${p^2} - 2q$

Answer 1

First of all, consider the value of the given determinant as $m$ and then expand it by the first row. Use an algebraic formula to simplify the determinant. Further find out the sum, product and the sum of two roots at a time of the given equation to get the required answer.

Complete step-by-step answer:
Given that $\alpha ,\beta ,\gamma$ are the roots of the equation ${x^3} + px + q = 0$.
Now, consider the value of the determinant as $m$. So, we have

\alpha &\beta &\gamma \\\ \beta &\gamma &\alpha \\\ \gamma &\alpha &\beta \end{array}} \right|$$ Opening the determinant by first row, we have

\Rightarrow m = \alpha \left[ {\gamma \beta - {\alpha ^2}} \right] - \beta \left[ {{\beta ^2} - \alpha \gamma } \right] + \gamma \left[ {\beta \alpha - {\gamma ^2}} \right] \\
\Rightarrow m = \alpha \beta \gamma - {\alpha ^3} + \alpha \beta \gamma - {\beta ^3} + \alpha \beta \gamma - {\gamma ^3} \\
\Rightarrow m = 3\alpha \beta \gamma - \left( {{\alpha ^3} + {\beta ^3} + {\gamma ^3}} \right)...................................\left( 1 \right) \\

We know that for a cubic equation $$a{x^3} + b{x^2} + cx + d = 0$$ the sum of the roots is given by $$\dfrac{{ - b}}{a}$$, the sum of two roots at a time is given by $$\dfrac{c}{a}$$ and the product of the roots are given by $$\dfrac{{ - d}}{a}$$. Now, for the given equation $${x^3} + px + q = 0$$ we have Sum of the roots is$$\alpha + \beta + \gamma = \dfrac{{ - 0}}{1} = 0...........\left( 2 \right)$$ Sum of two roots at a time is $$\alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{p}{1} = p..............\left( 3 \right)$$ Product of the roots is $$\alpha \beta \gamma = \dfrac{{ - q}}{1} = - q...........\left( 4 \right)$$ We know that

{\left( {a + b + c} \right)^3} = {a^3} + {b^3} + {c^3} + 3abc\left( {ab + bc + ca} \right) \\
{\left( {a + b + c} \right)^3} = {a^3} + {b^3} + {c^3} + 3\left[ {\left( {a + b + c} \right)\left( {ab + bc + ca} \right) - abc} \right] \\

By using this formula equation $$\left( 1 \right)$$ can be written as

\Rightarrow m = 3\alpha \beta \gamma - \left( {{\alpha ^3} + {\beta ^3} + {\gamma ^3}} \right) \\
\Rightarrow m = 3\alpha \beta \gamma - \left[ {{{\left( {\alpha + \beta + \gamma } \right)}^3} - 3\left\{ {\left( {\alpha + \beta + \gamma } \right)\left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right) - \alpha \beta \gamma } \right\}} \right]...........\left( 5 \right) \\

By substituting $$\left( 2 \right),\left( 3 \right),\left( 4 \right)$$ in equation$$\left( 5 \right)$$, we get

\Rightarrow m = 3\left( { - q} \right) - \left[ {{{\left( 0 \right)}^3} - 3\left\{ {\left( 0 \right)\left( p \right) - \left( q \right)} \right\}} \right] \\
\Rightarrow m = 3\left( { - q} \right) - \left[ {0 - 3\left( {0 - q} \right)} \right] \\
\Rightarrow m = - 3q - \left[ {0 - 3q} \right] \\
\Rightarrow m = - 3q + 3q \\
\therefore m = 0 \\

Therefore, the value of the given determinant is 0. **So, the correct answer is “Option B”.** **Note:** For a cubic equation $$a{x^3} + b{x^2} + cx + d = 0$$ the sum of the roots is given by $$\dfrac{{ - b}}{a}$$, the sum of two roots at a time is given by $$\dfrac{c}{a}$$ and the product of the roots are given by $$\dfrac{{ - d}}{a}$$. Always remember the algebraic formula $${\left( {a + b + c} \right)^3} = {a^3} + {b^3} + {c^3} + 3abc\left( {ab + bc + ca} \right)$$ to solve these kinds of problems.