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.