If $x + y + z = 17, xyz = 171$ and $xy + yz + 2zx = 111$,then the value of $\sqrt[3]{(x^{3}+y^{3}+z^{3}+xyz)}$ is:
Answer & explanation
Correct answer: option 4
x + y + z = 17
xy + yz + zx = 111
xyz = 171
We know that,
(x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx)
= 172 = x2 + y2 + z2 + 2 × 111
= x2 + y2 + z2 = 289 – 222
= x2 + y2 + z2 = 67
We also know that,
x3 + y3 + z3 – 3xyz = (x + y + z)[x2 + y2 + z2 – (xy + yz + zx)]
= x3 + y3 + z3 – 3 × 171 = 17 × (67 – 111)
= x3 + y3 + z3 – 513 = -748
= x3 + y3 + z3 = -748 + 513
= x3 + y3 + z3 = -235
$\sqrt[3]{(x^{3}+y^{3}+z^{3}+xyz)}$ = $\sqrt[3]{-235 +171)}$ = -4