If x + y + x = 5, 1/x+1/y+1/z=0 , xyz = 12 and x³ + y³ + z³ = 151, then the value of (x² + y² + z²) is:

If x + y + x = 5, 1/x+1/y+1/z=0 , xyz = 12 and x³ + y³ +…

If x + y + x = 5, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$ , xyz = 12 and x³ + y³ + z³ = 151, then the value of (x² + y² + z²) is:

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Correct answer: A

Explanation

The correct answer is option 1 i.e. 23.

$If x + y + z ≠ 0, (x^3 + y^3 + z^3) - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$

$(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$

⇒ $\frac{(yz + zx + xy)}{xyz}$ = 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

⇒ $\frac{115}{5}$ = x² + y² + z²

x² + y² + z² = 23