If x2 + y2 + z2 =133, xy +yz + zx = 114 and xyz = 216, then the value of x3 + y3 + z3 is:
Show answer & explanation
Correct answer: C
Explanation
Correct Answer is option 3 i.e. 1009
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Concept |
Understanding |
Application |
Fair calculation |
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Formulas: (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx) And x3 + y3 + z3 - 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx ) |
Given x2 + y2 + z2 = 133, xy + yz + zx = 114 and xyz = 216 |
(x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx) (x + y + z)2 = 133 + 2 × 114 (x + y + z)2 = 361 (x + y + z) = 19 Now x3 + y3 + z3 - 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx ) x3 + y3 + z3 – 3 × 216 = 19 × (133 – 114) x3 + y3 + z3 – 648 = 361 x3 + y3 + z3 = 1009 |
19 × (133 – 114) = 19 × 19 = 361 |