Express, express, express!

Purely algebraic solution (see what's below this for the step-by-step, complete solution):

$\frac {58a^3}{a^6 + a^5 + a^4 + a^3 + a^2 + a + 1}$

$= \frac {58a^3}{a^6 + a^5 + a^4 + a^3 + a^2 + a + \underline{3a - a^2}}$

$= \frac {58a^3}{a^6 + a^5 + a^4 + a^3 + 4a}$

$= \frac {58a^3}{a^6 + a^5 + a^4 + a^3 + 4(\underline{3a^2 - a^3})}$

$= \frac {58a^3}{a^6 + a^5 + a^4 - 3a^3 + 12a^2}$

$= \frac {58a^3}{a^6 + a^5 + a^4 - 3a^3 + 12(\underline{3a^3 - a^4})}$

$= \frac {58a^3}{a^6 + a^5 -11a^4 + 33a^3}$

$= \frac {58}{a^3 + a^2 -11a + 33}$

$= \frac {58}{\underline{3a^2 - a} + a^2 -11a + 33}$

$= \frac {58}{4a^2 -12a + 33}$

$= \frac {58}{4(\underline{3a - 1}) - 12a + 33}$

$=\frac {58}{29}$

$= \boxed{2}$

Step-by-step : This problem can be solved by rearranging the given equation and substituting it over and over in the algebraic expression.

$\frac {a}{a^2 + 1} = \frac {1}{3} \Rightarrow 3a = a^2 + 1$

Now that that's done, we can observe that the given algebraic expression, $\frac {58a^3}{a^6 + a^5 + a^4 + a^3 + a^2 + a + 1}$ , has a degree of six. In these kinds of problems it is a good idea to reduce the degree of the expression to arrive at a certain value (the constant).

To do that, it is a good idea to have the variables on one side of the fraction. Since $a^3$ is the only variable on the upper half of the equation, we seek to express the denominator in terms of powers greater than or equal to three .

We can rearrange our given equation for the constant in terms of the variables: $1 = 3a - a^2$ Substituting it into the algebraic expression, $\frac {58a^3}{a^6 + a^5 + a^4 + a^3 + a^2 + a + \underline{3a - a^2}} = \frac {58a^3}{a^6 + a^5 + a^4 + a^3 + 4a}$

Now, we repeat this process to raise the power. To do this for the variable $4a$ , we can multiply the given equation by $a$ to arrive at $a = 3a^2 - a^3$ . Substituting it into the algebraic expression,

$\frac {58a^3}{a^6 + a^5 + a^4 + a^3 + 4(\underline{3a^2 - a^3})} = \frac {58a^3}{a^6 + a^5 + a^4 - 3a^3 + 12a^2}$

Doing this again, $a^2 = 3a^3 - a^4$ .

Substituting, $\frac {58a^3}{a^6 + a^5 + a^4 - 3a^3 + 12(\underline{3a^3 - a^4})} = \frac {58a^3}{a^6 + a^5 -11a^4 + 33a^3}$

Now we can cancel $a^3$ .

$\frac {58a^3}{a^6 + a^5 -11a^4 + 33a^3} = \frac {58}{a^3 + a^2 -11a + 33}$

Having expressed all the variables on one side of the fraction, we can proceed to reduce the power of the expression. Looking back at our given equation expressed in degree 3, we can rearrange in terms of $a^3$ .

$a = 3a^2 - a^3 \Rightarrow a^3 = 3a^2 - a$

Substituting, $\frac {58}{\underline{3a^2 - a} + a^2 -11a + 33} = \frac {58}{4a^2 -12a + 33}$

Doing this again for $a^2$ , $1 = 3a - a^2 \Rightarrow a^2 = 3a - 1$ .

Substituting, $\frac {58}{4(\underline{3a - 1}) - 12a + 33} =\frac {58}{29} = \boxed{2}$