Question

If x, y, z are different

and$\left| \begin{matrix} x & x^{2} & 1 + x^{3} \\ y & y^{2} & 1 + y^{3} \\ z & z^{2} & 1 + z^{3} \end{matrix} \right|$ = 0; then xyz =

• A. –3
• B. –2
• C. –1
• D. None

Answer 1

Answer: (C)

–1

Answer 2

$\left| \begin{matrix} x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1 \end{matrix} \right|$ + $\left| \begin{matrix} x & x^{2} & x^{3} \\ y & y^{2} & y^{3} \\ z & z^{2} & z^{3} \end{matrix} \right|$ = 0

⇒ $\left| \begin{matrix} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{matrix} \right|$ + xyz $\left| \begin{matrix} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{matrix} \right|$ = 0

∴ xyz = –1.