Question

Verify that $x^3+y^3+z^3-3xyz=$ $\frac{1}{2}\left(x+y+z\right)[{(x-y)}^2+{(y-z)}^2+{(z-x)}^2]$

Answer 1

Verifying $x^3+y^3+z^3-3xyz=$$\frac{1}{2}\left(x+y+z\right)[{(x-y)}^2+{(y-z)}^2+{(z-x)}^2]$:

Taking R.H.S.,

Using the identity, ${\left(a-b\right)}^2=a^2-2ab+b^2$, we get

$$\begin{gathered} {\left(x-y\right)}^2=x^2-2xy+y^2 \\ {\left(y-z\right)}^2=y^2-2yz+z^2 \\ {\left(z-x\right)}^2=z^2-2zx+x^2 \end{gathered}$$

Now put these values in R.H.S., we get,

$\begin{array}{rcl}\frac{1}{2}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)[{(\mathrm{x}-\mathrm{y})}^2+{(\mathrm{y}-\mathrm{z})}^2+{(\mathrm{z}-\mathrm{x})}^2]& =& \frac{1}{2}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\times \left[\left({\mathrm{x}}^2-2\mathrm{xy}+{\mathrm{y}}^2\right)+\left({\mathrm{y}}^2-2\mathrm{yz}+{\mathrm{z}}^2\right)+\left({\mathrm{z}}^2-2\mathrm{zx}+{\mathrm{x}}^2\right)\right]\\ & =& \frac{1}{2}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\times \left[{\mathrm{x}}^2-2\mathrm{xy}+{\mathrm{y}}^2+{\mathrm{y}}^2-2\mathrm{yz}+{\mathrm{z}}^2+{\mathrm{z}}^2-2\mathrm{zx}+{\mathrm{x}}^2\right]\\ & =& \frac{1}{2}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\times \left[2{\mathrm{x}}^2-2\mathrm{xy}+2{\mathrm{y}}^2-2\mathrm{yz}+2{\mathrm{z}}^2-2\mathrm{zx}\right]\\ & =& \frac{1}{2}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\times 2\times \left[{\mathrm{x}}^2-\mathrm{xy}+{\mathrm{y}}^2-\mathrm{yz}+{\mathrm{z}}^2-\mathrm{zx}\right]\\ & =& \left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\times \left[{\mathrm{x}}^2-\mathrm{xy}+{\mathrm{y}}^2-\mathrm{yz}+{\mathrm{z}}^2-\mathrm{zx}\right]\\ & =& \mathrm{x}\times \left[{\mathrm{x}}^2-\mathrm{xy}+{\mathrm{y}}^2-\mathrm{yz}+{\mathrm{z}}^2-\mathrm{zx}\right]+\mathrm{y}\times \left[{\mathrm{x}}^2-\mathrm{xy}+{\mathrm{y}}^2-\mathrm{yz}+{\mathrm{z}}^2-\mathrm{zx}\right]+\mathrm{z}\times \left[{\mathrm{x}}^2-\mathrm{xy}+{\mathrm{y}}^2-\mathrm{yz}+{\mathrm{z}}^2-\mathrm{zx}\right]\\ & =& \left[{\mathrm{x}}^3-{\mathrm{x}}^2\mathrm{y}+{\mathrm{xy}}^2-\mathrm{xyz}+{\mathrm{xz}}^2-{\mathrm{zx}}^2\right]+\left[{\mathrm{x}}^2\mathrm{y}-{\mathrm{xy}}^2+{\mathrm{y}}^3-{\mathrm{y}}^2\mathrm{z}+{\mathrm{yz}}^2-\mathrm{zxy}\right]+\left[{\mathrm{zx}}^2-\mathrm{xyz}+{\mathrm{y}}^2\mathrm{z}-{\mathrm{yz}}^2+{\mathrm{z}}^3-{\mathrm{z}}^2\mathrm{x}\right]\\ & =& {\mathrm{x}}^3-\cancel{{\mathrm{x}}^2\mathrm{y}}+\cancel{{\mathrm{xy}}^2}-\mathrm{xyz}+\cancel{{\mathrm{xz}}^2}-\cancel{{\mathrm{zx}}^2}+\cancel{{\mathrm{x}}^2\mathrm{y}}-\cancel{{\mathrm{xy}}^2}+{\mathrm{y}}^3-\cancel{{\mathrm{y}}^2\mathrm{z}}+\cancel{{\mathrm{yz}}^2}-\mathrm{zxy}+\cancel{{\mathrm{zx}}^2}-\mathrm{xyz}+\cancel{{\mathrm{y}}^2\mathrm{z}}-\cancel{{\mathrm{yz}}^2}+{\mathrm{z}}^3-\cancel{{\mathrm{z}}^2\mathrm{x}}\\ & =& {\mathrm{x}}^3-\mathrm{xyz}+{\mathrm{y}}^3-\mathrm{zxy}-\mathrm{xyz}+{\mathrm{z}}^3\\ & =& {\mathrm{x}}^3+{\mathrm{y}}^3+{\mathrm{z}}^3-3\mathrm{xyz}\\ & =& \mathrm{L}.\mathrm{H}.\mathrm{S}\end{array}$

Hence, $x^3+y^3+z^3-3xyz=$$\frac{1}{2}\left(x+y+z\right)[{(x-y)}^2+{(y-z)}^2+{(z-x)}^2]$ is verified.