Question:
If $x+y+z=8$ and $x y+y z+z x=20$, Find the value of $x^{3}+y^{3}+z^{3}-3 x y z$
Solution:
Given,
x + y + z = 8 and xy + yz + zx = 20
We know that,
$(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(x y+y z+z x)$
$(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(20)$
$(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+40$
$8^{2}=x^{2}+y^{2}+z^{2}+40$
$64-40=x^{2}+y^{2}+z^{2}$
$x^{2}+y^{2}+z^{2}=24$
we know that, $x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)$
$\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right) x^{3}+y^{3}+z^{3}-3 x y z$
$=(x+y+z)\left[\left(x^{2}+y^{2}+z^{2}\right)-(x y+y z+z x)\right]$ here,
$x+y+z=8$
$x y+y z+z x=20$
$x^{2}+y^{2}+z^{2}=24$
$x^{3}+y^{3}+z^{3}-3 x y z=8[(24-20)]=8^{*} 4=32$
Hence, the value of $x^{3}+y^{3}+z^{3}-3 x y z$ is 32