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Question
If \[x^2 + \frac{1}{x^2} = 98\] ,find the value of \[x^3 + \frac{1}{x^3}\]
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Solution
In the given problem, we have to find the value of `x^3 + 1/x^3`
Given `x^3 + 1/x^3 = 98`
We shall use the identity `(x+y)^2 = x^2 + y^2 + 2xy`
Here putting `x^2 + 1/x^2 = 98`,
`(x+1/x)^2 = x^2 +1/x^2 + 2 xx x xx 1/x`
`(x+1/x)^2 = x^2 +1/x^2 + 2 xx x xx 1/x`
`(x+1/x)^2 = 98 + 2`
`(x+1/x)^2 = 100`
`(x+1/x) = sqrt100`
`(x+1/x) = ± 10`
In order to find `x^3 +1/x^3`we are using identity `a^3 +b^3 = (a+b)(a^2 +b^2 - ab)`
`x^3 + 1/x^3 = ( x+1/x) (x^2 + 1/x^2 - x xx 1/x)`
Here `(x+1/x) = 10` and `x^2 + 1/x^2 = 98`
`x^3 + 1 /x^3 = (x+1/x)(x^2 + 1/x^2 - x xx 1/x)`
` = 10 (98 - 1)`
` = 10 xx 97`
` = 970`
Hence the value of `x^3 + 1/x^3` is 970.
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