Question

If x + y + z = 19, xyz = 216 and xy + yz + zx = 114, then the value of $\sqrt{x^3+y^3+z^3+xyz}$ is.

• A. 32
• B. 28
• C. 30
• D. 35

Answer 1

Answer:

D. 35

Solution:

Given: 

x + y + z = 19

xyz = 216

xy + yz + xz = 114

Formula Used:

1.) (x + y + z)² = x² + y² + z² + 2(xy + yz + xz)

2.) x³ + y³ + z³ = 3(xyz) + (x + y +z ){(x² + y² + z² – (xy + yz + xz)}

Calculations:

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

⇒ (19)² = x² + y² + z² + 2(114)

⇒ x² + y² + z² = (19)² – 2(114)

⇒ x² + y² + z2 = 361 – 228 

⇒ x² + y² + z² = 133

Now,

x³ + y³ + z³ = 3(xyz) + (x + y +z ){(x² + y² + z² – (xy + yz + xz)}

⇒ x³ + y³ + z³ = 3(216) + (19)(133 – 114)

⇒ x³ + y³ + z³ = 648 + 19(19)

⇒ x³ + y³ + z³ = 648 + 361

⇒ x³ + y³ + z3 = 1,009

$\sqrt{x^3+y^3+z^3+xyz}$

$⇒ \sqrt{(1,009 + 216)}$

$⇒ \sqrt{1225}$ = 35

∴ The correct answer is 35