If $x+y+z=3, x y+y z+z x=-12$ and $x y z=-16$, then the value of $\sqrt{x^3+y^3+z^3+13}$ is:
Answer & explanation
Correct answer: option 3
Given,
x + y + z = 3, xy + yz + zx = -12 and xyz = -16
We know,
x3 + y3 + z3 – 3xyz = (x + y + z) [(x + y + z)2 – 3(xy + yz + zx)]
So,
x3 + y3 + z3 – 3xyz = (x + y + z) [(x + y + z)2 – 3(xy + yz + zx)]
= x3 + y3 + z3 – 3 (-16) = 3 [(3)2 – 3(-12)]
= x3 + y3 + z3 + 48 = 3(9 + 36)
= x3 + y3 + z3 + 48 = 3(45)
= x3 + y3 + z3 + 48 = 135
= x3 + y3 + z3 = 87
$\sqrt{x^3+y^3+z^3+13}$ = $\sqrt{87+13}$ = 10