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

Hint: 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
 \[ \Rightarrow m = \left| {\begin{array}{*{20}{c}}
  \alpha &\beta &\gamma \\
  \beta &\gamma &\alpha \\
  \gamma &\alpha &\beta
\end{array}} \right|\]
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.