sequence $2,10,18,26,$.... contains no cube I have came across the following problem:-
Show That The sequence $2,10,18,26,$.... contains no cube.

how can I able to solve this problem?can I get some help?
 A: Hint: work mod 8.  Look at elements of this sequence mod 8, and look at cubes mod 8.
A: Hint:Any cube number on divsion by 8 leaves a remainder of $0,1,3,5,7$ and the all the numbers of the given sequence is leaves a remainder of $2$ on division by $8$.
Reason :Any square number leaves a remainder of $1$ on division 8$((2k+1)^2=4k^2+4k+1=4k(k+1)+1=8c+1[\text{ as k(k+1) is even}])$ if the number is odd and else $0$ so cube leaves a remainder of $0,1,3,5,7$ on division by $8$. 
A: The sequence is $8n+2$.
Suppose $8n+2=m^3$.
Mod 8, the cubes are
1:1, 8:0, 27:3, 64:0, 125:5, 216:0, and 343:7.
None are congruent to 2, so 8n+2 is never a cube.
Alternative proof:
If $8n+2=m^3$,
$m$ must be even.
But then $m^3$ is divisible by 8,
and 8n+2 is not divisible by 8.
A: Clearly, the series is Arithmetic one with the common difference $=8,$ the first term being $2$
So, the $n$ the term will be $2+(n-1)8=8n-6$ which is even, so will be its cube root 
If the cube root is divisible by $2,$ the number $8n-6$ must be divisible by $2^3=8$
But $8n-6=2(4n-3)\implies $ the highest power of $2$ in $8n-6$ is exactly $1$ 
A: The terms are of the form- $8n+2=2(4n+1)$.So it has a factor 2 but is not divisible by 8.
