TITLE: Find the minimal $a+b$.
QUESTION [5 upvotes]: Let $x$ be real number such that $\frac{(1+x)^3}{1+x^3} = \frac{9}{13}$. If  $\frac{(1+x)^5}{1+x^5} = \frac{a}{b}$ where $a, b$  are positive integers. What is the minimal value of $a+b$.
My attempted work :
$\frac{(1+x^3)(1+x)^2}{(1+x)^5} = \frac{13}{9}$
$\frac{x^5+2x^4+x^3+x^2+2x+1}{(1+x)^5} = \frac{13}{9}$ ---[1]
Since $13(1+x)^3 = 9(1+x^3)$ 
$13x^3+39x^2+39x+13 = 9+9x^3$
$4x^3+39x^2+39x+4 = 0$ ---[2]
$(x+1)(4x^2+35x+4) = 0$
$(4x^2+35x+4) = 0$
$(x+1)^2 = -\frac{27}{4}x$
so $x = -\frac{4}{27}(x+1)^2$
[2] : multiplied through by $x$, we have 
$4x^4+2x^3+2x^2+4x = -37x^3 -37x^2$ ---[3]
[2] : $4x^3+4 = -39x^2-39x$
multiplied through by $\frac{37}{39}x$, we have 
$\frac{148}{39}x^4 + \frac{148}{39}x = -37x^3 -37x^2$ ---[4]
[3]=[4] : $4x^4+2x^3+2x^2+4x = \frac{148}{39}x^4 + \frac{148}{39}x = \frac{148}{39}x(x^3+1) = \frac{148}{39}x\left(\frac{13}{9}(1+x)^3\right) = \frac{148}{39}\left(-\frac{4}{27}(x+1)^2\right)\left(\frac{13}{9}(1+x)^3\right) = -\frac{148\cdot4\cdot13}{39\cdot27\cdot9}(x+1)^5 $
so $\frac{2x^4+x^3+x^2+2x}{(x+1)^5} = -\frac{148\cdot2\cdot13}{39\cdot27\cdot9}$ ---[5]
[1]-[5] : we have $\frac{1+x^5}{(1+x)^5} = \frac{13}{9} + \frac{148\cdot2\cdot13}{39\cdot27\cdot9}$
$\frac{(1+x)^5}{1+x^5} = \frac{729}{1349} = \frac{a}{b}$
$a+b = 729+1349= 2078$
Does minimal value in this problem mean gcd of numerator and denominator is equal to 1 ?
Is my answer correct ?
Is there an easier way to find the answer, please suggest ?
Thank you.

REPLY [2 votes]: EDIT: In short: yes, your answer is correct (you can ignore everything else here:)
$\frac{(1+x)^3}{1+x^3} = \frac{9}{13}$ implies $8 x = -35 \pm 3 \sqrt{129}$ from which you can compute $\frac{(1+x)^5}{1+x^5} = 729/1349$. So there is only one possible coprime pair $a,b$, making the minimality of $a+b$ an empty condition.
(Note: it may seem at first that you would get a cubic equation for $x$ but that's not true since the two $x^3$ terms cancel each other out. So all you have to do to find $x$ is to solve a quadratic equation. You did find this equation, at which point all that remained to be done is (a) solve it, and (b) substitute the result into $(1+x)^5/(1+x^5)$ which is not pleasant but not hard either).
PS. The quickest way to evaluate $(1+x)^5$ or $1+x^5$ at $x = $root of a quadratic equation $Q(x)$ is to divide those polynomials by $Q(x)$ with long division and to take the remainder.