Finding the First Decimal Digit!

Let us assume that $a= 2015^{2015}$ . Thus $N = \sqrt[3]{a^3 + 2a^2 + a}$ .

Let us have a general expression $\sqrt[3]{n^3 + 2n^2 + n}$ .

We can see that by putting $n=2, 3, 4, 5, 6, ...$ , in the above expression, the first decimal digit comes out to be $6$ . Therefore, we'll claim that it'll be $6$ for all integers $n$ , $n \geq 2$ and will try to prove it.

Now, We observe that: $n = \sqrt[3]{n^3} < \sqrt[3]{n^3 + 2n^2 + n} < \sqrt[3]{n^3 + 3n^2 + 3n + 1} = n+1$

It's sufficient to show that $\left(n + \dfrac{6}{10} \right) < \sqrt[3]{n^3 + 2n^2 + n} < \left( n + \dfrac{7}{10} \right) \quad ....(1)$

The first inequality of $(1)$ holds for $n \geq 2$ as: $(5n+3)^3 < 125(n^3 + 2n^2 +n) \Rightarrow 5n(5n-2) > 27$

The second inequality of $(1)$ holds for $n \geq 2$ as: $(10n+7)^3 > 1000(n^3 + 2n^2 +n) \Rightarrow 1000n^2 + 470n + 343 > 0$

Thus, $(1)$ holds for all $n \geq 2$ . Thus it holds for $a = 2015^{2015}$ , and so, the first decimal digit of $N$ is $\boxed{6}$ .