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! 
 A: Find $$  \left( c + d \sqrt 2 \right)^3 = 99 + 70 \sqrt 2 $$ and 
$$  \left( c - d \sqrt 2 \right)^3 = 99 - 70 \sqrt 2 $$ 
There is a Pell equation after that, as
$$ 99^2 - 2 \cdot 70^2 = 1. $$ In the language of real quadratic fields, this says $c^2 - 2 d^2 = 1$ as well. Indeed, the first one with integers is $c=3, d=2,$ and that works.
If you don't like fields, you still can use
$$ \left( c^2 - 2 d^2 \right)^3 = \left( c + d \sqrt 2 \right)^3 \left( c - d \sqrt 2 \right)^3 = \left(99 + 70 \sqrt 2 \right) \left(99 + 70 \sqrt 2 \right) = 1.  $$
A: Hint...try dividing both equations by $d^3$, and then making a substitution $x=\frac cd$. 
This gives $$x^3+6x=\frac{99}{d^3}$$ and $$3x^2+2=\frac{70}{d^3}$$
Eliminating the terms involving $\frac {1}{d^3}$ you ontain the polynomial $$70x^3-297x^2+420x-198=0$$ whose only real root is $x=\frac 32$ which by substitution leads to $c=3, d=2$. But I don't know if this is the best way...
A: \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 $70=-70$.
