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If x + y + z = 1, xy + yz + zx = – 1 and xyz = -1, then find the value of x3 + y3 + z3 |
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Answer» We know that x3 + y3 + z3 – 3 xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx) => x3 + y3 + z3 – 3xyz => (x + y + z) (x2 + y2 + z2 + 2xy + 2yz + 2zx – 3xy – 3yz – 3zx) (Subtracting and adding 2xy + 2yz + 2zx) => x3 + y3 + z3 – 3xyz = (x + y + z) {(x + y + z)2 – 3(xy + yz + zx)} => x3 + y3 + z3 – 3 x (-1) = 1 x {(1) 2 – 3 x (-1)} [Putting x + y + z = 1; xy + yz + zx = -1; xyz = -1] => x3 + y3 + z3 + 3 = 4 => x3 + y3 + z3 = 4 – 3 x3 + y3 + z3 = 1 |
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