Plenty of a

For ease of writing, we find the reciprocal, $\Sigma=a^3+a^2+a+1+\frac{1}{a}+\frac{1}{a^2}+\frac{1}{a^3}$ . We are told that $a^2+1=3a$ , or, $a+\frac{1}{a}=3$ (I). Squaring both sides of (I) gives $a^2+2+\frac{1}{a^2}=9$ , or $a^2+\frac{1}{a^2}=7$ . Cubing both sides of (I) gives $a^3+3a+\frac{3}{a}+\frac{1}{a^3}=27$ , or $a^3+\frac{1}{a^3}=18$ . Adding it all up, we find that $\Sigma=18+7+3+1=29$ , and our answer is $\boxed{\frac{1}{29}}$ .

From the first equation we have that $a^{2} + 1 = 3a \Longrightarrow a^{2} + a + 1 = 4a.$

So $S = \dfrac{a^{3}}{a^{6} + a^{5} + a^{4} + a^{3} + a^{2} + a + 1} = \dfrac{a^{3}}{a^{6} + a^{5} + a^{4} + a^{3} + 4a} =$

$\dfrac{a^{2}}{a^{5} + a^{4} + a^{3} + a^{2} + 4}.$

Now $a^{5} + a^{4} + a^{3} = a^{3}(a^{2} + a + 1) = a^{3}(4a) = 4a^{4}$ and

$a^{2} + 4 = (a^{2} + 1) + 3 = 3a + 3.$ So we now have that

$S = \dfrac{a^{2}}{4a^{4} + 3a + 3}.$

Now $a^{4} = (3a - 1)^{2} = 9a^{2} - 6a + 1 = 9(3a - 1) - 6a + 1 = 21a - 8,$ and so

$4a^{4} + 3a + 3 = 4(21a - 8) + 3a + 3 = 87a - 29 = 29(3a - 1) = 29a^{2}.$

So finally $S = \dfrac{a^{2}}{29a^{2}} = \boxed{\dfrac{1}{29}}.$

$\dfrac{a }{a^2+1}=\dfrac 1 3 ~ \implies~a^2=3a-1 ..(1)~\therefore~a^3=3a^2-a=8a-3..(2)\\From~(1)~~a=3-\dfrac 1 a ,~~\therefore \dfrac 1 a =3-a..(3)~~\therefore~\dfrac{ 1}{ a^2}=\dfrac 3 a -1=8-3a ..(4)\\ So~\dfrac {1}{a^3}=\dfrac 8 a -3=21-8a ..(5)\\\dfrac {a^3}{a^6+a^5+a^4+a^3+a^2+a+1} = \dfrac {1}{a^3+a^2+a+1+a^{-1}+a^{-2}+a^{-3}}\\Using~(1)~to~(5), a^3+a^2+a+1+a^{-1}+a^{-2}+a^{-3}\\=a^3+a^{-3}+a^2+a^{-2}+a+1+a^{-1}\\=8a-3+\color{#D61F06}{21}-8a+3a-1+\color{#D61F06}{8}-3a+a+1-a+3=29.\\\therefore~\dfrac {a^3}{a^6+a^5+a^4+a^3+a^2+a+1} =\dfrac 1 {29}\\~~\\Just~~another~~approach.$

A brute force approach would be to show that $a = \frac{3+\sqrt{5}}{2}$ from the 1st equation then substitute into the 2nd

a^3=(a^5(a+1)+a^3(a+1)+a(a+1)+1)×?=(a+1)×(a^5+a^3+a)+1=(a+1)×(a^5+a×(a^2+1))+1,from given a^2+1=3a,above exp =(a+1)×(a^5+3a^2+)1=(a+1)×a×(a^4+3a)+1=a×(a+1)×((a^2)2+3a)+1=a×(a+1)×((3a-1)^2+3a)+1=a×(a+1)×(9a^2-6a+1+3a)+1=a×(a+1)×(9a^2-3a+1)+1=a×(a+1)×(9×(3a-1)-3a)+1=a×(a+1)×(27a+-9-3a+1)+1=a×(a+1)×(24a-8)+1=a×(a+1)×8×(3a-1)+1=8a×(a+1)×a^2+1=(8a^3×(a+1)+1)×?=a^3 ,so ?=(8×(a+1)+a^-3 )^-1,a^2-3a+1=0,a=1.5+.5√5,substitute in expression by using calculator ?=1÷29######

brute force method
a=(3+sqrt(5))/2 or a=(3 - sqrt(5))/2
substitute a into (2) by MS-Excel
solution is 0.034483 =1/29

Taking the denominator of the latter rational expression, we write the following:

a^6 + a^5 + a^4 + a^3 + a^2 + a + 1 = (a^6 + 3a^4 + 3a^2 + 1) - 2a^4 - 2a^2 + (a^5 + a^3 + a)

or (a^2 + 1)^3 - (2a^2)(a^2 + 1) + a^5 + 2a^3 - a^3 + a;

or (a^2 + 1)^3 - (2a^2)(a^2 + 1) + a(a^2 + 1)^2 - a^3 (i).

Substitution of a^2 + 1 = 3a into (i) will now produce:

a^3 / [(3a)^3 - (2a^2) (3a) + a (3a)^2 - a^3] = a^3 / (27 - 6 + 9 - 1)*a^3 = 1/29.

Excellent problem!!!

For the first equation:

\­[ \large \frac a{a^2+1} = frac 1 3]

we can know that

\­(\a approx 2.618 )

So using that value we can solve the second equation

It is given that \­(\frac { { a }^{ 2 }+1 }{ a } =3\­) (just reciprocated the given condition)\­(\longrightarrow I\­)

Cubing both sides, we get \­(\frac { { a }^{ 6 }+3{ a }^{ 4 }+3{ a }^{ 2 }+1 }{ { a }^{ 3 } } =27\­)

\­(\Rightarrow \frac { { a }^{ 6 }+{ 3a }^{ 2 }({ a }^{ 2 }+1)+1 }{ { a }^{ 3 } } =27\­)

\­(\Rightarrow \frac { { a }^{ 6 }+a({ { a }^{ 2 }+1 })^{ 2 }+1 }{ { a }^{ 3 } } =27\­) using equation \­(I\­)

\­(\Rightarrow \frac { { a }^{ 6 }+{ a }^{ 5 }+2{ a }^{ 3 }+a+1 }{ { a }^{ 3 } } =27\­)

\­(\Rightarrow \frac { { a }^{ 6 }+{ a }^{ 5 }+{ a }^{ 3 }+a+1 }{ { a }^{ 3 } } =26\­)

Add \­(\frac { { a }^{ 2 }+1 }{ a }\­) to the LHS and add \­(3\­) to the RHS (Since they are equal)

\­(\Rightarrow \frac { { a }^{ 6 }+{ a }^{ 5 }+{ a }^{ 4 }+{ a }^{ 3 }+{ a }^{ 2 }+a+1 }{ { a }^{ 3 } } =29\­)

Hence required answer is the reciprocal =\­(\boxed { \frac { 1 }{ 29 } }\­)