2014/105) show that all terms of $\dfrac{107811}{3}$, $\dfrac{110778111}{3}$ and $\dfrac{111077781111}{3}$ ... are pefect cubes

let us look at 1st few terms

1st term = $\dfrac{107811}{3} = \dfrac{1 * 10^5 + 77 * 10^2 + 111}{3}$ 
2nd term = $\dfrac{110778111}{3} = \dfrac{11 * 10^7 + 777 * 10* 3 + 1111}{3}$
3rd term = $\dfrac{111077781111}{3} = \dfrac{111 * 10^9 + 7777 * 10^4 + 11111}{3}$

so nth term = $\dfrac{(10^n- 1) * 10^{2n+3} + 7 * 10^{n+2} -1)* 10^{n+1} + (10^{n+2} - 1)}{3*9}$
=$\dfrac{ (10^n-1) * 1000 * 10^{2n} + 7 *( 100 * 10^n - 1)(10 * 10^n) + (100 * 10^n - 1)}{27}$
=$\dfrac{1000 * 10^{3n} - 1000 * 10^{2n} + 7 *1000 * 10^{2n} - 7 *10 * 10 ^n + 100 * 10^n - 1}{27}$
= $\dfrac{1000 * 10^3n - 300 * 10 ^2n + 30 * 10 ^n -1}{27}$
= $\dfrac{10^{3n+3} - 3 * 10^{2n+2} + 3 * 10^{n+1} -1}{27}
= \dfrac{(10^{n+1} - 1)^3}{27}$

so nth term = $(\dfrac{10^{n+1} -1}{3})^3$

which is a perfect cube