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+z33xyz=(x+y+z)(x2+y2+z2xyzyzx)x^{3} + y^{3} + z^{3} - 3 x y z = \left(x + y + z\right) \left(x^{2} + y^{2} + z^{2} - x y - z y - z x\right) is an identity.

MathBot Answer:

The left-hand side of the equation evaluates to:

x3+y3+z33xyz=x33xyz+y3+z3\begin{align*}& x^{3} + y^{3} + z^{3} - 3 x y z\\=& x^{3} - 3 x y z + y^{3} + z^{3}\\\end{align*}

The right-hand side of the equation evaluates to:

(x+y+z)(x2+y2+z2xyzyzx)=(x+y+z)(x2+y2+z2xyyzzx)=(x+y+z)(x2+y2+z2xyyzxz)=x33xyz+y3+z3\begin{align*}& \left(x + y + z\right) \left(x^{2} + y^{2} + z^{2} - x y - z y - z x\right)\\=& \left(x + y + z\right) \left(x^{2} + y^{2} + z^{2} - x y - y z - z x\right)\\=& \left(x + y + z\right) \left(x^{2} + y^{2} + z^{2} - x y - y z - x z\right)\\=& x^{3} - 3 x y z + y^{3} + z^{3}\\\end{align*}

Since x33xyz+y3+z3=x33xyz+y3+z3x^{3} - 3 x y z + y^{3} + z^{3} = x^{3} - 3 x y z + y^{3} + z^{3}, the equation x3+y3+z33xyz=(x+y+z)(x2+y2+z2xyzyzx)x^{3} + y^{3} + z^{3} - 3 x y z = \left(x + y + z\right) \left(x^{2} + y^{2} + z^{2} - x y - z y - z x\right) is an identity.