We live in 2014!

The problem gives us a³-a²-a¹-a⁰=2014. We can rewrite it as a³-a²-a-2015=0. The natural solution to this equation must be a multiple of 2015, which is divisible by 5,13 and 31. The number that satisfies the equation is 13, so... 13.

a^3 - a² - & - 2015 = 0 then use the rational root theorem to find that 13 is a root for this equation . It can be written as ( a - 13) ( a² + 12a + 155) . The only solution is a = 13 :)

$a^{3}-a^{2}-a-1=2014$
$a$ must be > 10 since $10^{3}=1000$ and $a$ < 15 since $15^{3}=3375$
trying for $a$ =11 to 14, we know that 13 was the answer.
and also, if you notice that $a^{3}-a^{2}-a=2015$ , the last digit is 5, without doing calculation ( I mean simple ), we can directly know that 13 was the answer since $3^{3}-3^{2}-3=15$ also has 5 as the last digit

13^3-13^2-13^1-13^0=2014 I got it......

its 13 obvious every no can be obtain preceding its no