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Question
Find the cube of the following binomials expression :
\[\frac{3}{x} - \frac{2}{x^2}\]
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Solution
Given `(3/x-2/x^2)^3`
We shall use the identity `(a-b^3 ) = a^3-b^3- 3ab(a-b)`
Here `a=3/x,b = 2/x^2`
By applying the identity we get
`(3/x-2/x^2)^3 = (3/x)^3 - (2/x^2)^3 -3 (3/x)(2/x^2)(3/x-2/x^2)`
`= 27/x^3 - 8/x^6 -3 xx3/x xx 2/x^2 (3/x - 2/x^2)`
`= 27/x^3 - 8/x^6 -18/x^3(3/x - 2/x^2)`
`= 27/x^3 - 8/x^6 -(18/x^3 xx3/x) -(18/x^3xx2/x^2)`
`= 27/x^3 - 8/x^6 -(54/x^4 +36/x^5)`
`= 27/x^3 - 8/x^6 -54/x^4 +36/x^5`
Hence cube of the binomial expression of `(3/x-2/x^2)` is `= 27/x^3 - 8/x^6 -54/x^4 +36/x^5`.
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