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Question
Find the cube of the following binomials expression :
\[\frac{1}{x} + \frac{y}{3}\]
Solution
In the given problem, we have to find cube of the binomial expressions
Given `(1/x+y/3)^3`
We shall use the identity `(a+b)^3 = a^3+b^3+3ab(a+b)`
Here `a=1/x ,b=y/3`
By applying the identity we get
`(1/x+y/x)^3 = (1/x)^3 + (y/3)^3+3 (1/x)(y/3)(1/x+y/3)`
` = 1/x^3 +y^3/27+3 xx 1/x xx y/3 (1/x +y/3)`
` = 1/x^3 +y^3/27+ y/x (1/x +y/3)`
` = 1/x^3 +y^3/27+ y/x xx 1/x +y/x xxy/3`
` = 1/x^3 +y^3/27+ y/x^2+y/(3x)`
Hence cube of the binomial expression `1/x+y/3` is `1/x^3 + y^3/27 +y/x^2+y^2/(3x)`
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