Question

\(\begin{vmatrix} z & x & y \\[0.3em] z^2 &x^2 & y^2 \\[0.3em] z^4& x^4 &y^4 \end{vmatrix}\) = \(\begin{vmatrix} x & y &z \\[0.3em] x^2 &y^2 & z^2 \\[0.3em] x^4& y^4 &z^4 \end{vmatrix}\) = \(\begin{vmatrix} x^2 & y^2 & z^2 \\[0.3em] x^4 &y^4 & z^4 \\[0.3em] x& y&z \end{vmatrix}\) = xyz(x - y)(y - z)(z - x)(x + y + z)

Answer 1

Considering 

On taking xyz common

= – xyz(x – y) (z – y) [z² + y² + zy – x² – y² – xy]

= – xyz(x – y) (z – y) [(z – x) (z + x0 + y (z – x)]

= – xyz(x – y) (z – y) (z – x) (x + y + z)

= R.H.S

Hence proved.