If $x + y + z = 19, xyz = 216$ and $xy + yz + zx = 114$, then the value of $ x^{3}+ y^{3}+z^{3} + xyz$ is:
Answer & explanation
Correct answer: option 1
Given,
x + y + z = 19
xy + yz + zx = 114
(x + y + z)2 = (19)2
x2 + y2 + z2 + 2(xy + yz + zx) = 361
= x2 + y2 + z2 = 361 − 2(xy + yz + zx)
= x2 + y2 + z2 = 361 – 2 × 114 = 133
x2 + y2 + z2 = 133
We know , x3 + y3 + z3 − 3xyz = (x + y + z)( x2 + y2 + z2 − xy − yz − zx)
x3 + y3 + z3 − 3xyz = 19 × [133 – 114]
Adding 4xyz both sides,
x3 + y3 + z3 + xyz = 19 × 19 + 4 × 216 = 1225