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Let $\{a_n\}$ be the refered sequence,with $n$ a natural number,then, its general term is defined by: $a_n=\underbrace{111...11}_{n\ times}0\underbrace{777...77}_{n\ times}8\underbrace{111...11}_{ n+1\ times}: 3$ $\Rightarrow a_n=\frac{1.(10^{3n+2}+10^{3n+1}...+10^{2n+3})+7.(10^{2n+1}...+10^{n+2})+8.10^{n+1}+1.(10^n+10^{n-1}...+1)}{3}$ We have in the numerator three sums of three geometry sequence with ration $10$,such that, two have $n$ terms and the other $n+1$, in consequence: $a_n=\frac{10^{2n+3}(1-10^n)+7.10^{n+2}(1-10^n)-72.10^{n+1}+1.(1-10^{n+1})}{-9.3}$ $\Rightarrow a_n=\frac{-10^{3(n+1)}+3.10^{2(n+1)}-3.10^{n+1}+1}{-3^3}$ $\Rightarrow a_n=\frac{-(10^{n+1}-1)(10^{2(n+1)}+10^{n+1}+1)+3.10^{n+1}(10^{n+1}-1)}{-3^3}$ $\Rightarrow a_n={(\frac{10^{n+1}-1}{3})}^3$,it mean that all sequence term can be representated as a exact cube.

We let $t_n$ be the $n$th term of the sequence shown to get $t_n=\frac{\overbrace{1\dots1}^n0\overbrace{7\dots7}^n8\overbrace{1\dots1}^{n+1}}{3}$ We know that $10^n-1=\overbrace{9...9}^{n}$ so we can rewrite $t_n$ as $t_n=\frac{\frac{10^n-1}{9}10^{2n+3}+\frac{7(10^n-1)}{9}10^{n+2}+8\cdot10^{n+1}+\frac{10^{n+1}-1}{9}}{3}$ We can get rid of the $9$s to get $t_n=\frac{10^{3n+3}-10^{2n+3}+7\cdot10^{2n+2}-7\cdot10^{n+2}+72\cdot10^{n+1}+10^{n+1}-1}{27}$ We can combine like terms to get $t_n=\frac{1000\cdot10^{3n}-300\cdot10^{2n}+30\cdot10^n-1}{27}$ We can regroup powers of $10$ to get $t_n=\frac{10^{3n+3}-3\cdot10^{2n+2}+3\cdot10^{n+1}-1}{27}$ We take the cube root because we need to prove that $t_n$ is a perfect cube so $t_n=(\frac{10^{n+1}-1}{3})^3$ We need to prove that $\frac{10^{n+1}-1}{3}$ is an integer to make sure that its cube is a perfect cube. We know that $10^{n+1}-1\equiv 1^{n+1}-1\equiv 1-1\equiv 0\pmod3$ so $\frac{10^{n+1}-1}{3}$ is an integer and $t_n=(\frac{10^{n+1}-1}{3})^3$ is a perfect cube.