If x + y + z = 9 and xy + yz + zx = 23, then the value of (x3 + y3 + z3 – 3xyz) = ?
It is given that,
x + y + z = 9
And, xy + yz + zx = 23
As we know that,
(x + y + z)2 = (x2 + y2 + z2 + 2xy + 2yz + 2zx)
∴ (9)2 = [x2 + y2 + z2 + 2 (xy + yz + zx)]
x2 + y2 + z2 = 81 – 2 × 23
x2 + y2 + z2 = 81 – 46
x2 + y2 + z2 = 35
We also know that:
(x3 + y3 + z3 – 3xyz) = (x + y + z) [x2 + y2 + z2 – (xy + yz + zx)]
= 9 (35 – 23)
= 9 × 12
= 108
Hence, option A is correct