If x + y = 4, xy = 2, y + z = 5, yz = 3, z + x = 6 and zx = 4, then find the value of x3 + y3 + z3 – 3xy.
Show answer & explanation
Correct answer: C
Explanation
The correct answer is option 3 i.e 153.75
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Understanding |
Application |
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Given values: Adding first three equations to get (x+y+z) (x + y)2 + (y + z)2 + (z + x)2 = x2 + y2 + 2xy + y2 + z2 + 2yz + z2 + x2 + 2zxs x3 + y3 + z3 - 3xyz = (x + y + z) × (x2 + y2 + z2 - xy - yz - zx) |
x + y = 4 y + z = 5 z + x = 6 xy = 2 yz = 3 zx = 4 2(x + y + z) = 4 + 5 + 6 x + y + z = 7.5 (x + y)2 + (y + z)2 + (z + x)2 = 2(x2 + y2 + z2) + 2(xy + yz + zx) x3 + y3 + z3 - 3xyz = 7.5(29.5 - 2 - 3 - 4) x3 + y3 + z3 - 3xyz = 7.5 × 20.5 x3 + y3 + z3 - 3xyz = 153.75 |