Question

If x + y + z = 9 and xy + yz + zx = 26, find the value of x3 + y3 + z3 – 3xyz.

Answer 1

Answer: 27

Justification:

-> (x3 + y3 + z3 - 3xyz)

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

= (x + y + z) [(x + y + z)2 - 3(xy + yz + zx)]

= 9 x (81 - 3 x 26)

= (9 x 3)

= 27

Answer 2

The value of x³ + y³ + z³ - 3xyz comes out to be equal to 27.

Given:

x + y + z = 9

xy + yz + zx = 26

To Find:

The value of x³ + y³ + z³ - 3xyz.

Solution:

x + y + z = 9 - Equation (i)

xy + yz + zx = 26 - Equation (ii)

→ Squaring the equation (i) on both sides:

∴ (x + y + z)² = 9²

∴ x²+ y² + z² + 2xy + 2yz + 2zx = 81

∴ x²+ y² + z² + 2(xy + yz + zx) = 81

∴ x²+ y² + z² + 2(26) = 81

∴ x²+ y² + z² + 52 = 81

∴ x²+ y² + z² = 29 - Equation.(iii)

→ As we know: x³ + y³ + z³ - 3xyz = (x + y + z)(x²+ y² + z² - xy - yz - zx)

∴ x³ + y³ + z³ - 3xyz = (9)(29 - (xy - yz - zx))

∴ x³ + y³ + z³ - 3xyz = (9)(29 - 26)

∴ x³ + y³ + z³ - 3xyz = (9)(3)

∴ x³ + y³ + z³ - 3xyz = 27

Therefore the value of x³ + y³ + z³ - 3xyz comes out to be equal to 27.

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