If x + y + z = 8 and xy + yz + zx = 20, Find the value of


If $x+y+z=8$ and $x y+y z+z x=20$, Find the value of $x^{3}+y^{3}+z^{3}-3 x y z$




x + y + z = 8 and xy + yz + zx = 20

We know that,

$(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(x y+y z+z x)$






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,


$x y+y z+z x=20$


$x^{3}+y^{3}+z^{3}-3 x y z=8[(24-20)]=8^{*} 4=32$

Hence, the value of $x^{3}+y^{3}+z^{3}-3 x y z$ is 32



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