Proving that $8(x+1)^3 \ge (n+1)^3(x+n)$ for $n \ge 3$ and $x \ge n^2$

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For $x \ge n^2, n \ge 3$, what is the most straight foward to show that:

$$8(x+1)^3 \ge (n+1)^3(x+n)$$

Is it sufficient to show that:

$$8(n^2+1)^3 \ge (n+1)^3)(n^2+n)$$

And then show that:

$f(x) = 8(x+1)^3 - (n+1)^3(x+n)$ is increasing for $n \ge 3, x\ge n^2$.

Is there a simpler way?

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I would suggest differentiating the function f. If that function has no zero points above x = 3 it is increasing. This gives f'(x) = 24(x+1)²-(n+1)² = 24(x+1)²-n²-2n-1>24(x+1)²-4n²>=24(x+1)²-4x>0 for x>1 and the worst case scenario x=n. For the second equations, note that on the left there is a polynomial of degree 6 and a polynomial of degree 5 on the right