prove that x^3+y^3+z^3-3xyz =(x+y+z)(x^2+y^2+z^2-xy-zy-zx)
asked by guest
on Jan 28, 2025 at 6:46 pm
You asked:
Prove or disprove that the equation x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−zy−zx) is an identity.
MathBot Answer:
The left-hand side of the equation evaluates to:
=x3+y3+z3−3xyzx3−3xyz+y3+z3
The right-hand side of the equation evaluates to:
===(x+y+z)(x2+y2+z2−xy−zy−zx)(x+y+z)(x2+y2+z2−xy−yz−zx)(x+y+z)(x2+y2+z2−xy−yz−xz)x3−3xyz+y3+z3
Since x3−3xyz+y3+z3=x3−3xyz+y3+z3, the equation x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−zy−zx)
is an identity.