Question

Prove that $\left|\begin{array}{ccc}1& 1& 1\\ x^2& y^2& z^2\\ x^3& y^3& z^3\end{array}\right|$ = $(x-y)(y-z)(z-x)(xy+yz+zx)$

Answer 1

$\left|\begin{array}{ccc}1& 1& 1\\ x^2& y^2& z^2\\ x^3& y^3& z^3\end{array}\right|$

Performing column operation $C_2\to C_1-C_2$ and $C_3\to C_3-C_1$

we get-

$\left|\begin{array}{ccc}1& 0& 0\\ x^2& y^2-x^2& z^2-x^2\\ x^3& y^3-x^3& z^3-x^3\end{array}\right|$

$=\left|\begin{array}{ccc}1& 0& 0\\ x^2& (y-x)(y+x)& (z-x)(z+x)\\ x^3& (y-x)(y^2+x^2-yx)& (z-x)(z^2+z^2-zx)\end{array}\right|$

$=(y-x)(z-x)\left|\begin{array}{ccc}1& 0& 0\\ x^2& (y+x)& (z+x)\\ x^3& (y^2+x^2-yx)& (z^2+x^2-zx)\end{array}\right|$

$=(y-x)(z-x)\left\{\right(y+x\left)\right(z^3+x^2-zx)-(z+x\left)\right(y^2+x^2-yx\left)\right\}$

$=(x-y)(y-z)(z-x)(xy+yz+zx)$