Solving a system of equations: I am having problems solving a system of equations I haven't run into a system quite like this before and I am looking for some help, thanks!
$$a(1+\sqrt3)^2+b(1-\sqrt3)^2=3$$ $$a(1+\sqrt3)^3+b(1-\sqrt3)^3=8$$
How can I solve for $a$ and $b$?
 A: Write $z = 1+\sqrt3$ and $w= 1-\sqrt3$ then we have:
$$az^2+bw^2=3$$ $$az^3+bw^3=8$$ If we multiply first with $z$ and substrat from the second we get:
$$ b = {3z-8\over w(z^2-w^2)}  = {-5+3\sqrt{3}\over (1-\sqrt3)4\sqrt{3}}={3\sqrt{3}-5\over 4(\sqrt3-3)}= {2\sqrt3-3\over 12}$$  
Clearly $a$ is a conjugate of $b$ so $$a={2\sqrt3+3\over 12}$$ 
A: Guide:
Try solve for $a$ and $b$, where $A,B,C,D$ are constants.
$$aA+bB=3$$
$$aC+bD=8$$
This is just a regular simultaneous linear equations. 
If you are familiar with matrices, you might like to take inverse of a particular matrix.
A: Multiply the equations by $2(1+\sqrt{3})^2$ and $(1+\sqrt{3})^3$ and add them to get:
$$a(1+\sqrt{3})^4(2+(1+\sqrt{3})^2)=2(1+\sqrt{3})^2(3+4(1+\sqrt{3})) \Rightarrow $$
$$a=\frac{2(2+\sqrt{3})^2}{4(2+\sqrt{3})(3+\sqrt{3})}=\frac{3+\sqrt{3}}{12}.$$
Similarly (by multiplying and adding) you can find:
$$b=\frac{3-\sqrt{3}}{12}.$$
A: Write it as a system $Ax=v$ with $v=(3,8)^T$, $x=(a,b)^T$ and
$$
A=\begin{pmatrix} (1+\sqrt{3})^2 & (1-\sqrt{3})^2 \cr
(1+\sqrt{3})^3 & (1-\sqrt{3})^3\end{pmatrix}   
$$
Since $\det(A)=-8\sqrt{3}\neq 0$, the inverse $A^{-1}$ exists, so the unique solution is 
$$
\begin{pmatrix} a \cr b \end{pmatrix}=x=A^{-1}v=\frac{1}{4\sqrt{3}}\begin{pmatrix} 9-5\sqrt{3} \cr 15+9\sqrt{3} \end{pmatrix}
$$.
