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If x + y + z = 1, x2 + y2 + z2 = 2 and x3 + y3 + z3 = 3, then what is the value of xyz?
- A1/3
- B1/6
- C1/2
- D1/4
Answer not set
Discussion 1
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CCommunity
(x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx)
Putting the given values,
⇒ 2(xy + yz + zx) = 1 – 2 = -1
⇒ (xy + yz + zx) = -1/2
x3 + y3 + z3 – 3xyz = (x + y + z) × (x2 + y2 + z2 – xy – yz – zx)
Putting the known values,
⇒ 3 – 3xyz = 1 × [2 – (-1/2)]
⇒ 3xyz = 1/2
∴ xyz = 1/6
Putting the given values,
⇒ 2(xy + yz + zx) = 1 – 2 = -1
⇒ (xy + yz + zx) = -1/2
x3 + y3 + z3 – 3xyz = (x + y + z) × (x2 + y2 + z2 – xy – yz – zx)
Putting the known values,
⇒ 3 – 3xyz = 1 × [2 – (-1/2)]
⇒ 3xyz = 1/2
∴ xyz = 1/6