Question

can anyone please solve 13th question

Answer 1

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

C₂ ⇒ C₂-C₁ and C₃⇒C₃-C₁

$= \left[\begin{array}{ccc}1&0&0\\x&(y-x)&(z-x)\\x^3&(y^3-x^3)&(z^3-x^3)\end{array}\right]$

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

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

Expanding along R₁, we get

$= (y-x)(z-x)(z^2+zx+x^2 - y^2-yx-x^2)$

$= (y-x)(z-x)[z^2-y^2 +x(z-y)]$

$= (y-x)(z-x)[(z+y)(z-y)+x(z-y)]$

$= (y-x)(z-x)(z-y)(x+y+z)$

$= (x-y)(y-z)(z-x)(x+y+z)$

HOPE THIS HELPS!!

Answer 2

Answer:

Please see the attachment