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If x + y + z = 12, x^(2) + Y^(2) + z^(2)...

If `x + y + z = 12, x^(2) + Y^(2) + z^(2) = 96` and `(1)/(x)+(1)/(y)+(1)/(z)= 36` . Then find the value `x^(3) + y^(3)+z^(3).`

Text Solution

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To get the value of `x^(3)+Y^(3) + z^(3) - 3xyz`
`= (x + y + z ) (x^(2)+Y^(2)+x^(2) - xy- yz - zx)` …(1)
We need the value of xyz and xy + yz + zx.
We have `(x + Y + z)^(2) = 144`
`therefore x^(2)+y^(2)+z^(2)+2xy+2yz+2xz=144`
`rArr `96 + 2(xy + yz + xz) =144
`rArr `xy+ yz+ zx = 24
Given that `(1)/(x)+(1)/(x)+(1)/(z)= 36`
`therefore (xy+yz+zx)/(xyz)=36`
`rArr xyz = (24)/(36)=(2)/(3)`
From (1),
`x^(3)+y^(3)+z^(3) - 2 = 12xx(96-24)`
=864
So, `x^(3)+y^(3)+z^(3) = 866`
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