Given that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{5}$ and $abc = 5$, solve for $a^3 + b^3 + c^3$ I recently encountered this question and have been stuck for a while. Any help would be appreciated!
Q: Given that 
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{5} \tag{1} \label{eq:1}$$
$$abc = 5 \tag{2} \label{eq:2}$$
Find $a^3 + b^3 + c^3$. It wasn't specified in the question but I think it can be assumed that $a, b, c$ are real numbers.
My approach:
$$ ab + ac + bc = \frac{1}{5} abc = 1 $$
$$ a^3 + b^3 + c^3 = (a+b+c)^3 - 3[(a + b + c)(ab + ac + bc) - abc] $$
$$ a^3 + b^3 + c^3 = (a+b+c)^3 - 3(a+b+c) + 15 $$
From there, I'm not sure how to go about solving for $a + b + c$.
Something else I tried was letting $x = \frac{1}{a}, y = \frac{1}{b}, z = \frac{1}{c}$, so we get $$ xyz = x + y + z = \frac{1}{5} $$Similarly, I'm not sure how to continue from there. 

Let $k$ be an arbitrary real constant and $a$, $b$, $c$ be the roots of the cubic equation $x^3-kx^2+x-5=0$. Then we have $ab+bc+ca=1$ and $abc=5$. Also, we have $a+b+c=k$.
$$a^3+b^3+c^3=k(a^2+b^2+c^2)-(a+b+c)+15=k(a+b+c)^2-2k(ab+bc+ca)-(a+b+c)+15=k^3-2k-k+15=k^3-3k+15$$
Since there are infinitely many possible $k$ such that $x^3-kx^2+x-5=0$ has 3 real roots, the problem has infinitely many possible answers.
