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Question: The value of the determinant \(\left| \begin{matrix} 0 & b^{3} - a^{3} & c^{3} - a^{3} \\ a^{3} - b^...

The value of the determinant ∣0b3−a3c3−a3a3−b30c3−b3a3−c3b3−c30∣\left| \begin{matrix} 0 & b^{3} - a^{3} & c^{3} - a^{3} \\ a^{3} - b^{3} & 0 & c^{3} - b^{3} \\ a^{3} - c^{3} & b^{3} - c^{3} & 0 \end{matrix} \right| is equal to.

A

a3+b3+c3a^{3} + b^{3} + c^{3}

B

a3−b3−c3a^{3} - b^{3} - c^{3}

C

0

D

−a3+b3+c3- a^{3} + b^{3} + c^{3}

Answer

0

Explanation

Solution

∣0b3−a3c3−a3a3−b30c3−b3a3−c3b3−c30∣\left| \begin{matrix} 0 & b^{3} - a^{3} & c^{3} - a^{3} \\ a^{3} - b^{3} & 0 & c^{3} - b^{3} \\ a^{3} - c^{3} & b^{3} - c^{3} & 0 \end{matrix} \right|

0 & 1 & 1 \\ a^{3} - b^{3} & 1 & 1 \\ a^{3} - c^{3} & 1 & 1 \end{matrix} \right| = 0$$ $\lbrack C_{2} \rightarrow C_{2} - C_{1}$ and $C_{3} \rightarrow C_{3} - C_{1}\rbrack$ and then taking out common $(b^{2} - a^{3})$ from II<sup>nd</sup> column and ( $c^{3} - a^{3}$) from III<sup>rd</sup> column\].