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Question
Find the cube of the following binomials expression :
\[4 - \frac{1}{3x}\]
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Solution
Give `(4- 1/(3x))^3`
We shall use the identity `a^3- b^3 = a^3-b^3 - 3ab(a-b)`
Here `a=4,b=1/(3x)`
By applying in identity we get
`(4- 1/(3x))^3 = (4)^3 - (1/(3x))^3 - 3(4) (1/(3x)) (4-1/(3x))`
`= 4 xx 4xx 4 - (1 xx 1xx1)/(3x xx 3x xx 3x) - 12/(3x) (4-1/(3x))`
` = 64 - 1/(27x^3) - 4/x (4- 1/(3x))`
` = 64 - 1/(27x^3) - (4/x xx 4)-(4/x xx 1/(3x))`
` = 64 - 1/27x^3 - (16/x - 4/(3x^2))`
` = 64 - 1/27x^3 - 16/x + 4/(3x^2)`
Hence cube of the binomial expression of `(4- 1/(3x))^3` is `64 - 1/(27x^3) - 16/x + 4/(3x^2).`
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