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

જ.બા. $=\frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]$

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

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

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

$=2 \times \frac{1}{2} \times( x + y + z )\left( x ^{2}+ y ^{2}+ z ^{2}- xy - yz - zx \right)_{\text {0 }}^{\prime}$

$=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)=x^{3}+y^{3}+z^{3} u+3 x y z=$ ડા.બા.

$\therefore$ જ.બા. $=$ ડા.બા.