if `4a-\frac{4}{a}+3 = 0` , then the value of `a^3 - \frac {1}{a^3}+3 = ?` asked in { SSC CGL TIER

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 ----

$(a-\frac{1}{a})^3=(\frac{-3}{4})^3$

$a^3-\frac{1}{a^3} - 3 a \times\frac{1}{a}(a-\frac{1}{a})\times=\frac{-27}{64}$

$a^3-\frac{1}{a^3}- 3\times\frac{-3}{4} =\frac{-27}{64}$

$a^3-\frac{1}{a^3}+ \frac{9}{4}= \frac{-27}{64}$

$a^3-\frac{1}{a^3}= \frac{-27}{64}-\frac{9}{4}$

$a^3-\frac{1}{a^3}= \frac{-171}{64}$

now

$a^3-\frac{1}{a^3}+3 = \frac{-171}{64}+3$

$a^3-\frac{1}{a^3}+3 = \frac{21}{64}$

if $4a-\frac{4}{a}+3 = 0$ , then the value of $a^3 - \frac {1}{a^3}+3 = ?$ asked in { SSC CGL TIER - 1 2015 }