Using factor property of determinants prove that $$\begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} = (x-y)(y-z)(z-x)$$
Step-by-Step Solution
Key Concept: General
On putting $x = y, y = z, z = x$ we are getting value of determinant is 0 so by factor theorems $(x - y), (y - z)$ and $(z - x)$ are factors. So by factor theorem<br>$$\begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} = \lambda(x-y)(y-z)(z-x)$$<br>For $\lambda$ put $x = 0, y = 1, z = 2$<br>We get $\lambda = 1$<br>So R.H.S. $= (x - y)(y - z) (z - x)$
Correct Answer: A