Let $S$ be the set of ordered triples $(x,y,z)$ of real numbers for which
\[\log_{10}(x+y) = z \text{ and } \log_{10}(x^{2}+y^{2}) = z+1.\]
There are real numbers $a$ and $b$ such that for all ordered triples $(x,y.z)$ in $S$ we have $x^{3}+y^{3}=a \cdot 10^{3z} + b \cdot 10^{2z}.$ What is the value of $a+b?$
$\textbf{(A)}\ \frac {15}{2} \qquad  \textbf{(B)}\ \frac {29}{2} \qquad  \textbf{(C)}\ 15 \qquad  \textbf{(D)}\ \frac {39}{2} \qquad  \textbf{(E)}\ 24$
Let $x + y = s$ and $x^2 + y^2 = t$.   
Then, $\log(s)=z$ implies $\log(10s) = z+1= \log(t)$,so $t=10s$.   
Therefore, $x^3 + y^3 = s\cdot\dfrac{3t-s^2}{2} = s(15s-\dfrac{s^2}{2})$.  
Since $s = 10^z$, we find that $x^3 + y^3 = 15\cdot10^{2z} - (1/2)\cdot10^{3z}$.  
Thus, $a+b = \frac{29}{2}$ $\Rightarrow$ $\boxed{B}$