Evaluate a big expression. If:$$\lambda=\int_{0}^{1}\frac{dx}{1+x^3} ;$$
Then evaluate : $$ p=\lim_{n \rightarrow \infty}\left( \frac{\prod_{r=1}^{n}\left(n^3+r^3\right)}{n^{3n}} \right)^{1/n} ;$$
EDIT: I have now ended upto $$lnp=\int_{0}^{1} \ln(1+t^3).dt$$
Now, we have to evaluate this using $\lambda$ , howsoever .
 A: First, to evaluate $\lambda$, use partial fractions:
$$\frac{1}{1+x^3} = \frac{1}{3} \left ( \frac{1}{1+x} - \frac{x-2}{x^2-x+1} \right )$$
The second piece in the parentheses is more amenable to integration when expressed as follows:
$$\frac{x-2}{x^2-x+1} = \frac{x-1/2}{(x-1/2)^2+3/4} - \frac{3/2}{(x-1/2)^2+3/4}$$
Now, let 
$$A = \int_0^1 \frac{dx}{1+x} = \log{2}$$
$$B = \int_0^1 dx \: \frac{x-1/2}{(x-1/2)^2+3/4} = \left[\log{\left[ \left(  x-\frac{1}{2}\right)^2+\frac{3}{4}\right]}\right]_0^1 = 0$$
$$C = \int_0^1 dx \: \frac{3/2}{(x-1/2)^2+3/4} = \sqrt{3} \left[\arctan{\left [ \frac{2}{\sqrt{3}} \left( x-\frac{1}{2}\right)\right]}\right]_0^1 = \frac{\pi}{\sqrt{3}}$$
Therefore
$$\lambda = \frac{1}{3} (A - B + C) = \frac{1}{3} \log{2} + \frac{\pi}{3 \sqrt{3}}$$
Now for the limit, which is evaluated by taking logs of both sides:
$$\log{p} = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^n \log{\left(1+\frac{r^3}{n^3}\right)}$$
This is a Riemann sum, in the sense that
$$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^n f \left(\frac{r}{n}\right) = \int_0^1 dx \: f(x)$$
Therefore
$$\begin{align}\log{p} &= \int_0^1 dx \: \log{(1+x^3)}\\ &= [x \log{(1+x^3)}]_0^1 - \int_0^1 dx \: \frac{3 x^3}{1+x^3}\\ &= \log{2} - 3 \int_0^1 dx \: \left ( 1 - \frac{1}{1+x^3}\right)\\ &= \log{2}-3 + 3 \lambda\\&= 2 \log{2}-3 + \frac{\pi}{\sqrt{3}}\\ \therefore p &= 4 e^{-(3 - \frac{\pi}{\sqrt{3}})}\end{align}$$
A: Hint:  Take the log of the second expression
You'll get a Riemann sum expression which you can then use your value of $\lambda$ to solve.
