20
If x + y + z = 3 and x2 + y2 + z2 = 9, then the value of x3 + y3 + z3 – 3 xyz is?
- A27
- B30
- C0
- D90
Answer not set
Discussion 1
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0
CCommunity
We refer to the standard identity:
x3 + y3 + z3 – 3 xyz = (x + y + z)(x2 + y2 + z2– xy – yz – zx)
Since we have
x + y + z = 3
⇒ (x + y + z)2 = x2 + y2 + z2+ 2(xy + yz + zx) = 9
Hence, we have
2(xy + yz + zx) = 0
⇒ xy + yz + zx = 0
∴ x3 + y3 + z3 – 3 xyz = (3)(9 - 0) = 27
x3 + y3 + z3 – 3 xyz = (x + y + z)(x2 + y2 + z2– xy – yz – zx)
Since we have
x + y + z = 3
⇒ (x + y + z)2 = x2 + y2 + z2+ 2(xy + yz + zx) = 9
Hence, we have
2(xy + yz + zx) = 0
⇒ xy + yz + zx = 0
∴ x3 + y3 + z3 – 3 xyz = (3)(9 - 0) = 27