Correct Answer - Option 1 : 23
Given:
x + y + z = 5
\(\rm \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
xyz = 12
x3 + y3 + z3 = 151
Formulas used:
If x + y + z ≠ 0, (x3 + y3 + z3) - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - zx)
(x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx)
Calculation:
\(\rm \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
⇒ (yz + zx + xy)/xyz = 0
⇒ (xy + yz + zx) = 0 (1)
By using the formula, (x3 + y3 + z3) - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - zx)
⇒ 151 - 3 × 12 = 5 × (x2 + y2 + z2 - 0) ∵ (xy + yz + zx) = 0
⇒ 115/5 = x2 + y2 + z2
∴ x2 + y2 + z2 = 23
By using the formula, (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx)
⇒ (5)2 = x2 + y2 + z2 + 2 × 0 (From 1)
⇒ 25 - 0 = x2 + y2 + z2
∴ x2 + y2 + z2 = 25
Note: The data given in the question give two possible answers by using different identities.
Therefore care must be taken while choosing the correct option as RRB NTPC takes 25 too as the right answer.