$T(1) = 1 , T(n) = 2T(n/2) + n^3$? Divide and conquer $T(1) = 1 , T(n) = 2T(n/2) + n^3$? Divide and conquer, need help, I dont know how to solve it?

Hmm, possibly another way of heuristics is instructive.
First write the undisputable elements of the sequence:
$$\begin{array}  &
  a(1) &=&a(2^0) & = & 1 \\\
   a(2)&=&a(2^1) &=&10 &= & 2^3 + 2*1  &=& 2*(4^1+1) \\\
   a(4)&=&a(2^2) &=&84 &=& 4^3 + 2*10  &=& 4*(4^2+4^1+1)  \\\
   a(8)&=&a(2^3) &=&680 &=& 8^3 + 2*84 &=& 8^3+2*4^3+4*2^3+8*1^3\\\
       & &       & &                   &=& 8*(4^3+4^2+4^1+1)  \\\  
   \ldots &=&a(2^k)&=& \ldots 
   \end{array} $$
It is obvious how this can be continued, because at the exponent k we get always $8^k$ plus two times the previous, thus the weighted sum of all powers of 8 which can be expressed as consecutive powers of 4:
$$ a(2^k) = 2^k*(4^k+4^{k-1} \ldots +4^0)= 2^k*\frac{4^{k+1}-1}{4-1} $$
Now the step "divide" can be taken: the above gives also a meaningful possibility for interpolation of the non-explicitely defined elements. If we allow base-2 logarithms for k we get for
$$\begin{array} &
 & a(2^k) &= & 2^k*\frac{4^{k+1}-1}{4-1} \\\
  &      &= & 2^k*\frac{4*(2^{k})^2-1}{3} \\\
      \text{assuming }& k&=& \frac{\log(n)}{\log(2)} \\\
 & a(n)  &=& n*\frac{4*n^2-1}{3} \\\
  &      &=& n^3 + \frac{(n-1)n(n+1)}{3!} \\\
   &     &=& n^3 + 2*\binom{n+1}{3} \\\
 \end{array} $$
where the expression in the fourth line is the same as Fabian's result.
