Question

If $\alpha ,\beta ,\gamma$ are the cube roots of $8$ then find the value of \left| {\begin{array}{*{20}{c}} \alpha &\beta &\gamma \\\ \beta &\gamma &\alpha \\\ \gamma &\alpha &\beta \end{array}} \right| .
(A) $0$
(B) $1$
(C) $8$
(D) $2$

Answer 1

Hint : In this problem, first we will find the cube roots of the number $8$ . Then, we will find the sum of these roots. Then, we will evaluate the required value of determinant by using row-operations.

Complete step-by-step answer :
In this problem, it is given that $\alpha ,\beta ,\gamma$ are the cube roots of $8$ . Let us find the cube roots of $8$ . For this, we have to solve the equation $x = {\left( 8 \right)^{\dfrac{1}{3}}}$ . That is, we have to solve the equation ${x^3} - 8 = 0 \cdots \cdots \left( 1 \right)$ . We know that ${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)$ . Use this information to solve the equation $\left( 1 \right)$ . So, we can write
${x^3} - 8 = 0 \\\ \Rightarrow \left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right) = 0 \\\ \Rightarrow x - 2 = 0,\quad {x^2} + 2x + 4 = 0 \\\ \Rightarrow x = 2,\quad {x^2} + 2x + 4 = 0 \;$
Let us solve the second equation ${x^2} + 2x + 4 = 0$ by using the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ . Note that here $a = 1,b = 2,c = 4$ . Hence, we can write
${x^2} + 2x + 4 = 0 \\\ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt {4 - 16} }}{2} \\\ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt { - 12} }}{2} \\\ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt { - 1 \times 4 \times 3} }}{2} \\\ \Rightarrow x = \dfrac{{ - 2 \pm 2\sqrt 3 i}}{2}\quad \left[ {\because i = \sqrt { - 1} } \right] \\\ \Rightarrow x = - 1 \pm \sqrt 3 i \\\ \Rightarrow x = - 1 + \sqrt 3 i,\quad x = - 1 - \sqrt 3 i \;$
Hence, we have three roots $\alpha = 2,\quad \beta = - 1 + \sqrt 3 i,\quad \gamma = - 1 - \sqrt 3 i$ .
Let us find the sum of these three roots. So, we can write
$\alpha + \beta + \gamma = 2 + \left( { - 1 + \sqrt 3 i} \right) + \left( { - 1 - \sqrt 3 i} \right) = 0$
Now we are going to find the value of \left| {\begin{array}{*{20}{c}} \alpha &\beta &\gamma \\\ \beta &\gamma &\alpha \\\ \gamma &\alpha &\beta \end{array}} \right| by using row-operations.
Let us say D = \left| {\begin{array}{*{20}{c}} \alpha &\beta &\gamma \\\ \beta &\gamma &\alpha \\\ \gamma &\alpha &\beta \end{array}} \right| . Apply ${R_1} \to {R_1} + {R_2} + {R_3}$ . So, we can write
D = \left| {\begin{array}{*{20}{c}} {\alpha + \beta + \gamma }&{\beta + \gamma + \alpha }&{\gamma + \alpha + \beta } \\\ \beta &\gamma &\alpha \\\ \gamma &\alpha &\beta \end{array}} \right|
Taking $\alpha + \beta + \gamma$ common from the first row. So, we can write
D = \left( {\alpha + \beta + \gamma } \right)\left| {\begin{array}{*{20}{c}} 1&1&1 \\\ \beta &\gamma &\alpha \\\ \gamma &\alpha &\beta \end{array}} \right| \\\ \Rightarrow D = 0\left| {\begin{array}{*{20}{c}} 1&1&1 \\\ \beta &\gamma &\alpha \\\ \gamma &\alpha &\beta \end{array}} \right|\quad \left[ {\because \alpha + \beta + \gamma = 0} \right] \\\ \Rightarrow D = 0 \;
Hence, if $\alpha ,\beta ,\gamma$ are the cube roots of $8$ then value of \left| {\begin{array}{*{20}{c}} \alpha &\beta &\gamma \\\ \beta &\gamma &\alpha \\\ \gamma &\alpha &\beta \end{array}} \right| is zero.

Note : Sometimes it is difficult to evaluate the determinant directly. So, in that case we can use row-operations to evaluate the determinant. To find cube roots of any number $N$ , we have to solve the equation ${x^3} - N = 0$ . To solve the quadratic equation $a{x^2} + bx + c = 0$ , we can use the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ . In the given problem, we get one real root $x = 2$ and two complex roots $x = - 1 \pm \sqrt 3 i$ . Every cubic equation has at least one real root.