How show $8|[(\sqrt[3]{n}+\sqrt[3]{n+2})^3]+1$ let $n$ be postive integer numbers,and such $n>2$,show that

$$
8\,\,\left.\right\vert\,\,\left\lfloor%
\left(\vphantom{\Large A}\sqrt[3]{n\,} + \sqrt[3]{n + 2\,}\,\right)^{3}%
\right\rfloor + 1
$$
  where $\left\lfloor x\right\rfloor$ is is the largest integer not greater than $x$,

My try: 
since

$$(a+b)^3=a^3+b^3+3ab(a+b)$$
  then
  $$(\sqrt[3]{n}+\sqrt[3]{n+2})^3=n+n+2+3\sqrt[3]{n^2+2n}(\sqrt[3]{n}+\sqrt[3]{n+2})$$
  so
  $$[(\sqrt[3]{n}+\sqrt[3]{n+2})^3]=2n+2+[3\sqrt[3]{n^2+2n}(\sqrt[3]{n}+\sqrt[3]{n+2})]$$

then I can't,Thank you 
 A: Claim For $n >2$ we have
$$\left\lfloor%
\left(\vphantom{\Large A}\sqrt[3]{n\,} + \sqrt[3]{n + 2\,}\,\right)^{3}%
\right\rfloor =8n+7$$
This is equivalent to
$$8n+7 \leq 
\left(\vphantom{\Large A}\sqrt[3]{n\,} + \sqrt[3]{n + 2\,}\,\right)^{3}< 8n+8 \Leftrightarrow \\
 8n+7 \leq 2n+2+[3\sqrt[3]{n^2+2n}(\sqrt[3]{n}+\sqrt[3]{n+2})] < 8n+8 \Leftrightarrow \\
6n+5 \leq3\sqrt[3]{n^2+2n}(\sqrt[3]{n}+\sqrt[3]{n+2}) < 6n+6 $$ 
Hopefully this part is right: By Jensen
$$\sqrt[3]{n}+\sqrt[3]{n+2}< 2 \sqrt[3]{\frac{n+n+2}{2}}=2\sqrt[3]{n+1}$$
also
$$3\sqrt[3]{n^2+2n}<3\sqrt[3]{n^2+2n+1}=3\sqrt[3]{(n+1)^2}\,.$$
Multiplying these two inequalities yields the upperbound.
For the lowerbound we know 
$$3\sqrt[3]{n^2+2n}(\sqrt[3]{n}+\sqrt[3]{n+2}) =3\sqrt[3]{n^3+2n^2}+3\sqrt[3]{n^3+4n^2+4n}$$
Now, I think you can prove:
$$3\sqrt[3]{n^3+4n^2+4n} \geq 3n+3.5$$
and  for $n > 2$ we have
$$3\sqrt[3]{n^3+2n^2} > 3n+1.5$$
Much Simpler Solution
By AM-GM
$$\left( \sqrt[3]{n\,} + \sqrt[3]{n + 2\,} \right)^3 > 8 \sqrt{n^2+2n\,} >8n+7$$
with the last inequality being true for $n >3$. The case $n=3$ is trivial to check.
As Calvin pointed, by Jensen we have
$$\left(\vphantom{\Large A}\sqrt[3]{n\,} + \sqrt[3]{n + 2\,}\,\right)^{3} < \left(2\vphantom{\Large A}\sqrt[3]{n+1\,} \right)^{3} =8n+8$$
