Solving Symmetrical Equations Algebraically I'm doing some Cambridge STEP papers and have come across a tricky set of equations. 
\begin{align*} 99 &= c^3 + 6 cd^2  \tag{1} \\
70 &= 3c^2d + 2d^3 \tag{2} \end{align*} 
From looking around, I've found the easiest way to solve these for the real solutions is to note that $ c^3 < 99 $ and $ 2d^3 < 70 $ and then to work numerically. My question is: How can these be solved algebraically ?
Many thanks! 

\begin{gather}
99=c^3+6cd^2\\
d=\sqrt{\frac{99-c^3}{6c}}
\end{gather}
Substituting into the other equation yields
\begin{gather}
3c^2\sqrt{\frac{99-c^3}{6c}}+2\left(\frac{99-c^3}{6c}\right)^{\frac{3}{2}}=70\\
\sqrt{\frac{99-c^3}{6c}}\left(3c^2+\frac{198-2c^3}{6c}\right)=70\\
\sqrt{\frac{99-c^3}{6c}}\left(\frac{16c^3+198}{6c}\right)=70\\
\left(\frac{99-c^3}{6c}\right)\left(\frac{16c^3+198}{6c}\right)^2= 4900\\
(-c^3+99)(16c^3+198)^2-4900(6c)^3=0\\
-256c^9+19008c^6-470340c^3+3881196=0\\
\end{gather}
which is a rather tedious polynomial to solve. It can be factored using the rational roots theorem, but I will not do that here. Using a calculator gives me the only real root $c=3$. Substituting back into the expression for $d$ gives
$$d=\sqrt{\frac{99-27}{18}}=\sqrt{\frac{72}{18}}=\sqrt{4}=2$$
So a solution of this system of equations is $(c,d)=(3,2)$. $d=-2$ does not work because the second equation will yield $70=-70$.
