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

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

 

 

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