Find condition for $\alpha + \beta+ \gamma=\alpha . \beta. \gamma$ Find condition for real numbers $\alpha , \beta, \gamma$ such that $\alpha + \beta+ \gamma=\alpha . \beta. \gamma$
Solution:
we have this  set for example:
 $\sqrt 3+\sqrt 3+\sqrt 3=\sqrt 3.\sqrt 3.\sqrt 3$
We find a general relation which define the condition. We have:
$\alpha +\beta =- \gamma(1-\alpha . \beta)$
$\frac{\alpha + \beta}{(1-\alpha . \beta)}=-\gamma$
We compare this relation with $ tan( A+B)=\frac { tan A+tan B}{1-tan A.tan B}$
Let $\alpha=tan A$,$\beta=tan B$ and $\gamma =tan C$
So we may write:
$tan(A+B)=-tan C=tan (\pi - C)$
$A+B+C=\pi$
Or
$tan^{-1}\alpha + tan^{-1}\beta + tan^{-1}\gamma= \pi$
defines the condition.
Question:  Is this only solution?
 A: It might be useful to introduce complex numbers for the calculations. It gives basically the same answer, but it allows us to interpret the condition visually and it explains how we should deal with those pesky $\tan$'s.
Note that the imaginary part of $(1+\alpha i)(1+\beta i)(1+\gamma i)$ is equal to $\alpha+\beta+\gamma - \alpha\beta\gamma$. So the equation is equivalent to $\mathrm{im} (1+\alpha i)(1+\beta i)(1+\gamma i)=0$. A complex number is real iff its argument is zero or $\pi$. The argument of a product of complex numbers is equal to the sum of arguments of the complex numbers. Hence the condition becomes
$$ \arg(1+\alpha i) + \arg(1+\beta i) + \arg(1+\gamma i) = 0,$$
or
$$\mathrm{atan2}(\alpha)+\mathrm{atan2}(\beta)+\mathrm{atan2}(\gamma)=0.$$
The $\mathrm{atan}2$ is useful for calculating arguments of complex numbers. Basically, $\mathrm{atan}2$ takes in account in which quadrant the complex number lies. The definition of the $\mathrm{atan}2$ function in terms of the standard $\arctan$ can be found here.
