Question

18. If x + y + z =1. then the value of the
1
determinant
1 1 1
x y z
x³ y³ z³
(A) 2
(B) 3
(C) 1
(D) 0​

Answer 1

Given,

$\[x + y + z = 1\]$

$\[\begin{array}{*{20}{c}}\vline& 1&1&1\vline& \\\vline& x&y&z\vline& \\\vline& {{x^3}}&{{y^3}}&{{z^3}}\vline& \end{array}\]$

Solution,

Calculate the determinant.

$\[ = \begin{array}{*{20}{c}}\vline& 1&1&1\vline& \\\vline& x&y&z\vline& \\\vline& {{x^3}}&{{y^3}}&{{z^3}}\vline& \end{array}\]$

$\[{C_3} \to {C_3} - {C_2},{C_2} \to {C_2} - {C_1}\]$

$\[ = \begin{array}{*{20}{c}}\vline& 1&0&0\vline& \\\vline& x&{y - x}&{z - y}\vline& \\\vline& {{x^3}}&{{y^3} - {x^3}}&{{z^3} - {y^3}}\vline& \end{array}\]$

Expand it,

$\[ = 1 \times \left[ {\left( {y - x} \right) \times \left( {{z^3} - {y^3}} \right) - \left( {{y^3} - {x^3}} \right)\left( {z - y} \right)} \right]\]$

$\[ = \left( {y - x} \right) \times \left( {z - y} \right)\left( {{z^2} + {y^2} + zy} \right) - \left( {{y^2} + {x^2} + xy} \right)\left( {y - x} \right)\left( {z - y} \right)\]$

$\[ = \left( {y - x} \right) \times \left( {z - y} \right)\left[ {\left( {{z^2} + {y^2} + zy} \right) - \left( {{y^2} + {x^2} + xy} \right)} \right]\]$

$\[ = \left( {y - x} \right) \times \left( {z - y} \right)\left[ {{z^2} + {y^2} + zy - {y^2} - {x^2} - xy} \right]\]$

$\[ = \left( {y - x} \right) \times \left( {z - y} \right)\left[ {{z^2} - {x^2} + zy - xy} \right]\]$

$\[ = \left( {y - x} \right) \times \left( {z - y} \right)\left( {z - x} \right)\left[ {z + x + y} \right]\]$

Put $\[x + y + z = 1\]$

$\[ = \left( {y - x} \right) \times \left( {z - y} \right)\left( {z - x} \right)\]$

Hence the determinant is $\[\left( {y - x} \right) \times \left( {z - y} \right)\left( {z - x} \right)\]$.

No given option is matched with correct answer.