Question

x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx) prove.

Answer 1

here is u answer. just do it by simple multiplication

Answer 2

$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$, proved.

Step-by-step explanation:

To prove that, $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

R.H.S.$=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

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

$=(x^3+y^2x+z^2x-x^2y-xyz-zx^2)+(yx^2+y^3+yz^2-xy^2-y^2z-yzx)+(x^2z+y^2z+z^3-xyz-yz^2-z^2x)$

$=x^3+y^2x+z^2x-x^2y-xyz-zx^2+yx^2+y^3+yz^2-xy^2-y^2z-yzx+x^2z+y^2z+z^3-xyz-yz^2-z^2x$

$=x^3+y^3+z^3-xyz-xyz-xyz$

= L.H.S., it is proved.

Hence, $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$, proved.