We have $a^3+b^3$ and $ab$, how we can calculate $a+b$? One of my friends is a high school student, he asked me this question. It's soluble by use of General formula for cubic roots, because:  
$$(a+b)^3=a^3+b^3+3ab(a+b)$$
But he looked for a simple answer (e.g use of identities).  
Edit: Thanks to the Travis and Andre Nicolas comments, the original question said $a^3+b^3=\frac{7}{2}$ and $ab=.75$
 A: There's no way to improve on this, at least for general $a^3 + b^3$ and $ab$: If we relabel the given values and denote $x:=a+b$, we're solving $x^3 + p x + q = 0$, but this a general depressed cubic, which is the form to which one typically reduces a general cubic when solving the general cubic equation.
On the other hand, for the special values $a^3 + b^3 = \frac{7}{2}$ and $ab = \frac{3}{4}$ OP gives in the comments, the resulting equation is:
$$\tfrac{1}{4}(4 x^3 - 9 x - 14) = 0 .$$
The Rational Root Theorem gives finitely many possible rational roots, and checking them we find that $x = 2$ is a solution. Polynomial long division then gives the quadratic factor of the above cubic, but it turns out to have negative discriminant, so the other solutions are nonreal.
A: $$x_{start}=(a+b)$$
$$x_{n+1} =\sqrt{\frac{a^3+b^3}{x_n}+3ab }$$
This is not single shot operation but an iteration technique.
Start with a reasonable first iteration value of $x$. Substitute it in right hand side. Plug in the left hand side resulting value again into into right side. Repeat the iteration until input and output difference is within error limit required.
