If x + y + z = 3, and x2 + y2 + z2 = 101, then what is the value of √(x3 + y3 + z3 - 3xyz) ?
Show answer & explanation
Correct answer: A
Explanation
The correct answer is Option 1 i.e 21.
| Understanding/Application |
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x + y + z = 3 x2 + y2 + z2 = 101 x3 + y3 + z3 - 3xyz = (x + y + z)[(x + y + z)2 - 3(xy + yz + zx)] ....(1) (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx) |
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On putting the above values - 32 = 101 + 2(xy + yz + zx) (xy + yz + zx) = -92/2 (xy + yz + zx) = -46 |
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From eq.1 we get - x3 + y3 + z3 - 3xyz = (3)[(3)2 - 3(-46)] x3 + y3 + z3 - 3xyz = 441 √(x3 + y3 + z3 - 3xyz) = √441 = 21 |