if `4a-\frac{4}{a}+3 = 0` , then the value of `a^3 - \frac {1}{a^3}+3 = ?`
solution - `4a-\frac{4}{a}+3 = 0`
`4 ( a- \frac{1}{a} ) = -3`
`(a-\frac{1}{a}) =\frac{-3}{4}`
On taking power on 3 bothsides we have now ----
`(a-\frac{1}{a})^3=(\frac{-3}{4})^3`
`a^3-\frac{1}{a^3} - 3 a \times\frac{1}{a}(a-\frac{1}{a})\times=\frac{-27}{64}`
`a^3-\frac{1}{a^3}- 3\times\frac{-3}{4} =\frac{-27}{64}`
`a^3-\frac{1}{a^3}+ \frac{9}{4}= \frac{-27}{64}`
`a^3-\frac{1}{a^3}= \frac{-27}{64}-\frac{9}{4}`
`a^3-\frac{1}{a^3}= \frac{-27}{64}-\frac{9}{4}`
`a^3-\frac{1}{a^3}= \frac{-171}{64}`
now
`a^3-\frac{1}{a^3}+3 = \frac{-171}{64}+3`
`a^3-\frac{1}{a^3}+3 = \frac{21}{64}`
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Mathematics