a,b,c are positive integers forming an increasing G.P and b-a is a perfect cube and $\displaystyle log_6(a)$+$\displaystyle log_6(b)$+$\displaystyle log_6(c)$ =6 then a+b+c=?
$\displaystyle \begin{aligned}
\log_6\,a + \log_6\,b + \log_6\,c &= 6 \\
6^{\log_6\,a + \log_6\,b + \log_6\,c} &= 6^6 \\
6^{\log_6\,a} \cdot 6^{\log_6\,b} \cdot 6^{\log_6\,c} &= 6^6 \\
abc &= 6^6
\end{aligned}$
If a, b, and c form a geometric progression then b = ar and c = ar^2 (for some ratio r):
$\displaystyle \begin{aligned}
abc &= 6^6 \\
a(ar)(ar^2) &= 6^6 \\
a^3 r^3 &= 6^6 \\
ar &= 36 \\
b &= 36
\end{aligned}$
$\displaystyle b - a$ has to be a perfect cube, or $\displaystyle 36 - a$ has to be a perfect cube. There are a few possibilities for a (35, 28, 9), but the only one that will work in our case is if a = 9.
If a = 9 and b = 36, then r = 4, which makes c = 144. So a + b + c = 189.