If $\left| \begin{matrix} x & x^{2} & 1 + x^{3} \ y & y^{2} & 1 + y^{3} \ z & z^{2} & 1 + z^{3} \end{matrix} ight| = 0\$and$\ x,y,z\$are all distinct, then $xyz$ is equal to

Question

If $\left| \begin{matrix} x & x^{2} & 1 + x^{3} \ y & y^{2} & 1 + y^{3} \ z & z^{2} & 1 + z^{3} \end{matrix} ight| = 0\$and$\ x,y,z\$are all distinct, then $xyz$ is equal to

Dr. Abubacker · Accepted Answer

(a)
Given, $\left| \begin{matrix}
x & x^{2} & 1 + x^{3} \
y & y^{2} & 1 + y^{3} \
z & z^{2} & 1 + z^{3}
\end{matrix} ight| = 0$
$\Rightarrow \left| \begin{matrix}
x & x^{2} & 1 \
y & y^{2} & 1 \
z & z^{2} & 1
\end{matrix} ight| + \left| \begin{matrix}
x & x^{2} & x^{3} \
y & y^{2} & y^{3} \
z & z^{2} & z^{3}
\end{matrix} ight|$=0
$\Rightarrow \left| \begin{matrix}
x & x^{2} & 1 \
y & y^{2} & 1 \
z & z^{2} & 1
\end{matrix} ight| + xyz\left| \begin{matrix}
1 & x & x^{2} \
1 & y & y^{2} \
1 & z & z^{2}
\end{matrix} ight| = 0$
$\Rightarrow (1 + xyz)\left| \begin{matrix}
1 & x & x^{2} \
1 & y & y^{2} \
1 & z & z^{2}
\end{matrix} ight| = 0$
$\Rightarrow 1 + xyz = 0$
$\Rightarrow \ \ xyz = - 1$

Q. If $\left| \begin{matrix} x & x^{2} & 1 + x^{3} \\ y & y^{2} & 1 + y^{3} \\ z & z^{2} & 1 + z^{3} \end{matrix} \right| = 0\$and$\ x,y,z\$are all distinct, then $xyz$ is equal to

Explanation

Reviewed by Dr. Abubacker

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