If $x + y + z = 8$ and $xy + yz+ zx = 20$, find the value of $x^3 + y^3 + z^3 – 3xyz$.


Given: 

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

To do: 

We have to find the value of $x^3 + y^3 + z^3 – 3xyz$.

Solution: 

We know that,

$x^3 + y^3 + z^3 – 3xyz = (x + y + z) (x^2 + y^2 + z^2 -xy -yz - zx)$

$x + y + z = 8$

Squaring both sides, we get,

$(x + y + z)^2 = (8)^2$

$x^2 + y^2 + z^2 + 2(xy + yz + zx) = 64$

$x^2 + y^2 + z^2 + 2 \times 20 = 64$

$x^2 + y^2 + z^2 + 40 = 64$

$x^2 + y^2 + z^2 = 64 - 40 = 24$

Therefore,

$x^3 + y^3 + z^3 - 3xyz = (x + y + z) [x^2 + y^2 + z^2 - (xy + yz + zx)]$

$= 8(24 - 20)$

$= 8 \times 4$

$= 32$

Hence, $x^3 + y^3 + z^3 - 3xyz = 32$.

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Updated on: 10-Oct-2022

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