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Question
Find the cube of : `( 3a - 1/a ) (a ≠ 0 )`
Sum
Solution
( a - b )3 = a3 - 3a2b + 3ab2 - b3
`( 3a - 1/a )^3`
= `(3a)^3 - 3 × (3a)^2 × 1/a + 3. 3a (1/a)^2 - (1/a)^3`
= `27a^3 - 3 . 9a^2. 1/a + 9a. 1/a^2 - 1/a^3`
= `27a^3 - 27a + 9/a - 1/a^3`.
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