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. 
 A: As DanielV suspects, the question is indeed ill-posed: there are multiple possible outcomes for $a^3 + b^3 + c^3$. First of all notice that if $a$, $b$, and $c$ are positive, then in order for (1) to hold, each of them should be larger than 5. But this results in a contradiction for (2). It follows that at least one should be negative, but then (2) gives that at least two of $a,b,c$ should be negative. Without loss of generality, assume that $a > 0$ and $b, c < 0.$ 
Since $$\frac 1 b = \frac 1 5 - \frac 1 c - \frac 1 a,$$ we find 
$$\frac{5}{ac} = b = \frac{1}{\frac 1 5 - \frac 1 c - \frac 1 a} = \frac{5ac}{ac-5a-5c},$$ 
so that $ (ac)^2 = ac - 5a - 5c.$ Solving this quadratic equality for $c$ gives
$$c = \frac {a-5}{2a^2} \pm\frac{\sqrt{(5-a)^2 - 20a^3}}{2a^2}.$$
By symmetry the same equality should hold for $b$,
$$b = \frac {a-5}{2a^2} \pm\frac{\sqrt{(5-a)^2 - 20a^3}}{2a^2}.$$ 
Supposing that $b$ and $c$ are not both equal, we can take $b$ with the + sign and $c$ with the minus sign. For all values of $a$ for which the argument of the square root is non-negative, the above choices for $b$ and $c$ solve the stated system. It can be checked that different choices of $a$, and corresponding choices for $b$ and $c$, yield different outcomes for $a^3 + b^3 +c^3$.
We assumed above that $b \neq c$. We can also assume that $b = c$, which gives the equations $a b^2 = 5$ and $ \frac 1 a + \frac 2 b = \frac 1 5$, resulting in the cubic equality $-b^3 - 10 + b = 0$. This may be solved to  give (approximately) $b = c = -2.30891$ and $a = 0.937901$, yielding yet another value for $a^3 + b^3 + c^3$.
A: 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.
