Let $a,ar,a{r^2},a{r^3},a{r^4},a{r^5}$ be six terms of a $G.P.$ where 'a' is first term andr is  common ratio.

According to given condition, we have 

$a{r^3} – a = 5 \Rightarrow a\left( {{r^3} – 1} \right) = 52\,\,\,\,……\left( 1 \right)$

and $a + ar + a{r^2} = 26$

$ \Rightarrow a\left( {1 + r + {r^2}} \right) = 26\,\,\,\,\,\,……\left( 2 \right)$

To find : $a\left( {1 + r + {r^2} + {r^3} + {r^4} + {r^5}} \right)$

Consider

$a\left[ {1 + r + {r^2} + {r^3} + {r^4} + {r^5}} \right]$

$ = a\left[ {1 + r + {r^2} + {r^3}\left( {1 + r + {r^2}} \right)} \right]$

$ = a\left[ {1 + r + {r^2}} \right]\left[ {1 + {r^3}} \right]\,\,\,\,\,\,……\left( 3 \right)$

Divide $(1)$ by $(2)$, we get

$\frac{{{r^3} – 1}}{{1 + r + {r^2}}} = 2$

we know ${r^3} – 1 = \left( {1 + {r^3}} \right)\left( {1 + r + {r^2}} \right)$

$\therefore r – 1 = 2 \Rightarrow r = 3\,\,\,\,$  and $a = 2$

$\therefore a\left( {1 + r + {r^2} + {r^3} + {r^4} + {r^5}} \right)$

$ = a\left( {1 + r + {r^2}} \right)\left( {1 + {r^3}} \right)$

$ = 2\left( {1 + 3 + 9} \right)\left( {1 + 27} \right)$

$ = 26 \times 28 = 728$