2017/032) If $a,b,c$ are in H.P then prove that $a^3b^3 + b^3c^3 + c^3 a^3 = (9ac-6b^2)a^2c^2$

$a,b,c$ are in HP
hence $\frac{1}{a} + \frac{1}{c} =  \frac{2}{b}$
cube both sides to get  $\frac{1}{a^3} + \frac{1}{c^3} + 3 \frac{1}{a}  \frac{1}{c}(\frac{1}{a} + \frac{1}{c}) =  \frac{8}{b^3}$
or   $\frac{1}{a^3} + \frac{1}{c^3} + 3 \frac{1}{a}  \frac{1}{c}\frac{2}{b}  =  \frac{8}{b^3}$
or $\frac{1}{a^3} + \frac{1}{c^3} + \frac{1}{b^3}  =  \frac{9}{b^3} -  3 \frac{1}{a}  \frac{1}{c}\frac{2}{b}$
multiply both sides by $a^3b^3c^3$ to get
$b^3c^3 + a^3b^3 + a^3c^3 = 9a^3c^3 -  6 a^2b^2c^2 = (9aac-6b^2) a^2c^2$