What is a faster way to do this problem?
Solve the equation $$ \frac{1}{x^2+6x-15} + \frac{2}{x^2+6x+3} = \frac{3}{x^2+6x+1} $$
This is from a timed test, so the fastest answers would be the best. I know the old reliable way is just to multiply the denominators, but I'm pretty sure there is a faster way.
On
Rewrite it as
$$\frac1{(x+3)^2-24}+\frac2{(x+3)^2-6}=\frac3{(x+3)^2-8}$$
and notice that $\frac1{24}+\frac26=\frac9{24}=\frac38$. If you don’t notice that (and it really is worth trying $x=-3$, just on spec), you can let $y=x+3$, rewrite as
$$\frac1{y-24}+\frac2{y-6}=\frac3{y-8}$$
and solve fairly quickly:
$$\frac{3y-54}{y^2-30y+144}=\frac3{y-8}\,,$$
$$\frac{y^2-26y+144}{y^2-30y+144}=1\,,$$
and $y=0$.
On
If $y=x^2+6x+9$ (I chose $9$ because $x^2+6x+9$ is a perfect square), then your equation becomes$$\frac1{y-24}+\frac2{y-6}=\frac3{y-8}.\tag1$$But$$\frac1{y-24}+\frac2{y-6}-\frac3{y-8}=\frac{12y}{(y-24)(y-6)(y-8)}$$and therefore the only root of $(1)$ is $0$. So, since $x^2+6x+9=(x+3)^2$, the only root of your equation is $-3$.
The first thing that comes to my mind is to let $y = x^2 + 6x + 1$. Then $$\frac{1}{y-16} + \frac{2}{y+2} = \frac{3}{y}.$$ Then solve for $y$, and then solve the quadratic for $x$.