If \(x+y+z=2\) and \(xy+yz+zx=-11\), then find the value of \(x^3+y^3+z^3-3xyz\).
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 74
The identity \(x^3+y^3+z^3-3xyz=(x+y+z)\left(x^2+y^2+z^2-xy-yz-zx\right)\) needs \(x^2+y^2+z^2\), which comes from the square of the sum.
\(x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)=4-2(-11)=4+22=26\)
\(x^2+y^2+z^2-(xy+yz+zx)=26-(-11)=37\)
\(x^3+y^3+z^3-3xyz=2\times37=74\)