Can we solve for $a$ and $b$ only knowing the value of $z$? We have the following equation for $z$ > 0

$ a^2b+ab^2 = z $

Only the value of $z$ is known. How can we solve this equation to get the values of $a$ and $b$?
Simplifying,

$ab(a+b) = z $ 

eg. $z=84$ 
then $a = 3$, $b=4$ 
Any approach or suggestions or trial and error method ?
 A: If $a,b,z \in \mathbb{R}$ then there are an infinite number of solutions.
If we assume $a=b$ then we have $2a^3=z \Rightarrow a=b=\root 3 \of {\frac z 2}$.
But if we assume $2a=b$ then we have $6a^3=z \Rightarrow a=\root 3 \of {\frac z 6}, b=2\root 3 \of {\frac z 6}$.
And, in general, if $ka=b$ then $(k+k^2)a^3=z \Rightarrow a=\root 3 \of {\frac z {k+k^2}}, b=k\root 3 \of {\frac z {k+k^2}}$.
On the other hand if $a,b$ are constrained to be integers then you can have zero, one, two or an infinite number of solutions, depending on the value of $z$.
A: Let ${a+b\over2}=:m$. This implies $a+b=2m$ and $2m ab=z$. It then follows from Vieta's theorem that $a$ and $b$ are the roots of the quadratic equation
$$x^2-2mx+{z\over2m}=0\ .$$
This implies that
$$\{a,b\}=\left\{m-\sqrt{m^2-{z\over2m}}, \ m+\sqrt{m^2-{z\over2m}}\right\}\ .$$
If $m<0$ then $-{z\over2m}>0$. We therefore obtain two pairs $\{a,b\}$, whereby the numbers $a\ne0$, $b\ne0$ have different signs.
If $m>0$ we need $m^2\geq {z\over2m}$ in order to obtain real solutions. If $m=\root 3\of{z/2}$ we obtain $a=b=m$, and if $m>\root 3\of{z/2}$we obtain a pair $\{a,b\}$ of different positive numbers.
See the figure linked in Matti P.'s comment.
