[Algebra]
Problem: Suppose that $a(x+b(x+3))=2(x+6)$ for all values of $x$. Determine $a$ and $b$.

Solution: Since the given equation is true for all values of $x$, then it is true for any particular value of $x$ that we try.

If $x=-3$, the equation becomes $a(-3+b(0))=2(3)$ or $-3 a=6$ and so $a=-2$.

If $x=0$, the equation becomes $-2(0+b(3))=2(6)$ or $-6 b=12$ and so $b=-2$.

Therefore, $a=-2$ and $b=-2$.
We expand both sides of the equation:

$$
\begin{aligned}
a(x+b(x+3)) & =2(x+6) \\
a(x+b x+3 b) & =2 x+12 \\
a x+a b x+3 a b & =2 x+12 \\
(a+a b) x+3 a b & =2 x+12
\end{aligned}
$$

Since this equation is true for all values of $x$, then the coefficients on the left side and right side must be equal, so $a+a b=2$ and $3 a b=12$.

From the second equation, $a b=4$ so the first equation becomes $a+4=2$ or $a=-2$.

Since $a b=4$, then $-2 b=4$ and so $b=-2$.

Thus, $a=b=-2$.

Final answer: -2,-2