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Question
If x + y + z = 8 and xy +yz +zx = 20, find the value of x3 + y3 + z3 −3xyz
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Solution
n the given problem, we have to find value of x3 + y3 + z3 −3xyz
Given x + y + z = 8 , xy +yz +zx = 20
We shall use the identity
`(x+y+z)^2 = x^2 + y^2 + z^2 + 2 (xy + yz +za)`
`(x+y+z)^2 = x^2 + y^2 + z^2 +2 (20)`
`64 = x^2 + y^2 +z^2 + 40`
`64 - 40 = x^2 + y^2 + z^2`
`24 = x^2 + y^2 + z^2`
We know that
`x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz -zx)`
`x^3 + y^3 + z^3 - 3xyz = (x+y+z)[(x^2 + y^2 + z^2 )- (xy - yz -zx)]`
Here substituting `x+y +z = 8,xy +yz + zx = 20,x^2 +y^2 + z^2 = 24 ` we get
`x^3 + y^3 + z^3 -3xyz = 8 [(24 - 20)] `
` = 8 xx 4`
` =32`
Hence the value of x3 + y3 + z3 −3xyz is 32.
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