Prove that there no $k$ such that : $3n^{2}+3n+7=k^{3}$ Problem : 
Let $n,k$ be natural numbers 
Prove that there no $k$ such that : 

$$3n^{2}+3n+7=k^{3}$$ 

I see this short solution from a book : 
We use $\pmod{9}$ 
$k^{3}\equiv 0,1,8\pmod{9}$ 
And $3n^{2}+3n+7\equiv 4,7\pmod{9}$ 
$\implies $ no solution ? 
But I don't understand why he chose mod $9$ and how we know that $k^{3}\equiv 0,1,8\pmod{9}$ ? 
And how he know $3n^{2}+3n+7\equiv 4,7\pmod{9}$ ?
Can someone explain to me or give me another solution !

The reason for choosing 9 is that n is not divisible by 3 , let $n=3m+r$ , we can write:
$n^2=9m^2+6mr+r^2$
⇒ $$3n^2+3n+7 ≡(R=3r^2+3r+7) \ mod (9)$$
$r= 0, 1, 2$ 
$r=0$ ⇒ $R=7$ ⇒ $3n^2+3n+7 ≡7 \mod(9)$
$r=1$ ⇒ $R=13$ ⇒ $$3n^2+3n+7 ≡13 \mod(9)≡4 \mod(9)$$
$r=2$ ⇒ $R=7 \mod (9)$
⇒ $$3n^2+3n+7 ≡(4, 7) \mod(9)$$
$k≡(0, 1, 2, . . .8)\ mod (9)$
⇒ $K^3≡(0, 1, 8, 0, 1, 8, 0, 1, 8)\ mod (9) ≡(0, 1, 8)\ mod (9)$
