If $p(x)$ is a polynomial of degree 3 such that $p(i) = {1\over1+i}$ for all $a=\{1,2,3,4\}$. Then find $p(5)$. If $p(x)$ is a polynomial of degree $3$ such that $p(i)$ = $\frac{1}{1+i}$ for all $a=\{1,2,3,4\}$. Then find $p(5)$.

My attempt : (1) First obviously I thought of solving the four
  equations which can be generated by assuming polynomial as
  $ax^{3}+bx^{2}+cx+d$. But that would be a very lengthy solution. 
(2) In other attempt, I let
$g(x)=p(x) - \frac1{1+x}$ believing that some calculus concepts might be
  applicable since $g(x)=0$ at $1,2,3,4$ but I couldn't think of anything
  that works out.
Also, we "cannot" make $g(x) =k(x-1)(x-2)(x-3)(x-4)$ since we can only
  write that for polynomials since we can comment on their maximum
  number of real roots by looking at the degree. 
That rules out another possible method.

So, how can this question be solved?

I am writing this answer based on the hints provided by Daniel Fischer( in the comments section) so that others can also benefit from this answer.
Consider a polynomial function $h(x)=p(x)(x+1)-1$ 
$x=1,2,3,4$ are clearly the roots of this polynomial therefore $h(x)$ can also be written as
$h(x)=k(x-1)(x-2)(x-3)(x-4)$
So, $h(x)=p(x)(x+1)-1=k(x-1)(x-2)(x-3)(x-4)$----(1)
To find the value of k, we put $x=-1$ in (1)
( why? Because it would make p(x) vanish, so we wouldn't need more values of p(x) to find k)
Putting $x=-1$, $0-1=k(-2)(-3)(-4)(-5)$ 
This gives $k=\frac{-1}{120}$
Now to find $p(5)$,
we put $x=5$ in eq(1),  this gives the value of $p(5)=\frac{2}{15}$
