Question

If x + y + z = 1, xy + yz + zx = -1 and xyz = -1 then the value of \(\sqrt[3]{x^3+y^3+z^3}\) is:
1. 1
2. -1
3. 2
4. -2

Answer 1

Correct Answer - Option 1 : 1

Given:

x + y + z = 1, xy + yz + zx = -1 and xyz = -1

Calculation:

⇒ x³ + y³ + z³ - 3xyz = (x + y + z) (x² + y² + z² - xy - yz - zx)

⇒ x³ + y³ + z³ - 3 × (-1) = (1) × {x² + y² + z² - (-1)}

⇒ x³ + y³ + z³ + 3 = x² + y² + z² + 1     ..... (1)

⇒ (x + y + z)² = x² + y² + z² + 2xy+ 2yz + 2zx

⇒ x² + y² + z² = 3 ..... (2)

⇒ From equation (1) and (2)

⇒ x³ + y³ + z³ = 1

⇒ ∛(x3 + y3 + z3) = ∛1 = 1

∴ The required result will be 1.