Let $p^3+q^3=4$ and $pq=2/3$ . Find $p+q$. 
Let $p^3+q^3=4$ and $pq=\frac{2}{3}$ . Find $p+q$.

A graphing calculator can find values of $p$ and $q$ numerically.  As one can see from the graph below, the two solutions are approximately $(0.4, 1.6)$ and $(1.6, 0.4)$:

However, I am interested in a symbolic solution.  Is there a method to solve this problem quickly without having to use a graphing calculator?
 A: Hint:
$$p^3+q^3=4$$
$$p^3q^3=(\frac{2}{3})^3 \Rightarrow q^3=\frac{8}{27p^3}$$
The equation 
$$p^3+\frac{8}{27p^3}=4$$
is quadratic in $p^3$.
A: $$(p+q)^3=[p^3+q^3]+3pq(p+q)$$
$$(p+q)^3=[4]+3\frac{2}{3}(p+q)$$
$$(p+q)^3=[4]+2(p+q)$$
$$u^3-2u-4=0$$ where $u=p+q$
$$(u-2)(u^2+2u+2)=0$$
Therefore $u=2$, or some complex solutions.
Note that the root 2 could have been found by the rational root theorem.
The long way:
$$pq=\frac{2}{3}\implies q=\frac{2}{3p}$$
$$p^3+q^3=4$$
$$\implies p^3+\frac{8}{27p^3}=4$$
$$\implies p^6-4p^3+\frac{8}{27}=0$$
Let $u=p^3$
$$\implies u^2-4u+\frac{8}{27}=0$$
$$\implies u= 2\pm\frac{10\sqrt3}{9}$$
$$\implies p=\sqrt[3]{2\pm\frac{10\sqrt3}{9}}$$
$$\implies q=\frac{2}{3[\sqrt[3]{2\pm\frac{10\sqrt3}{9}}]}$$
$$\cdots$$
Can you take it from there?
A: we have $p^3 + q^3 = 4$ and $pq = 2/3$ so $p^3 q^3 = 8/27.$  therefore 
$$0= (x - p^3)(x-q^3) = x^2 - (p^3 + q^3)x + p^3q^3 = x^2 - 4x + 8/27$$ solving this quadratic equation we find 
$x = \dfrac{4 \pm \sqrt{16 - 4*8/27}}{2} =\dfrac{6\sqrt 3 \pm 10}{ 3\sqrt 3}$
so $$p+q = \left( \dfrac{6\sqrt 3 + 10}{ 3\sqrt 3}\right)^{1/3} + \left( \dfrac{6\sqrt 3 - 10}{ 3\sqrt 3}\right)^{1/3} $$
