If x + y + z = 19 & x2 + y2 + z2 = 133, then the value of x3 + y3 + z3 - 3xyz is:
Show answer & explanation
Correct answer: C
Explanation
The correct answer is Option 3 i.e. 361.
Given:
⇒ (x + y + z) = 19, (x2 + y2 + z2) = 133
Now,
⇒ (x + y + z) = 19
Squaring both sides:
⇒ (x + y + z)2 = 192
⇒ x2 + y2 + z2 + 2(xy + yz + zx) = 361
⇒ 2(xy + yz + zx) = 361 – 133
⇒ 2(xy + yz + zx) = 228
⇒ (xy + yz + zx) = 114
Now,
⇒ x3 + y3 + z3 - 3xyz
⇒ (x + y + z)(x2 + y2 + z2 - xy - yz - zx)
⇒ 19 × (133 – 114) = (19 × 19) = 361