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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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