Show that the equation $\sqrt{ax+\alpha}+\sqrt{bx+\beta}+\sqrt{cx+\gamma}=0$ reduces to a simple equation. I have got a question which looks like follow:

Show that the equation $\sqrt{ax+\alpha}+\sqrt{bx+\beta}+\sqrt{cx+\gamma}=0$ reduces to a simple equation if $\sqrt{a}\pm\sqrt{b}\pm\sqrt{c}=0$.

I am totally confused and don't even know from where should I start.
Side note: I don't know what is a simple equation (I think it is something which is not filthy like given equation).
Any hint/suggestion is heartily welcome. Thanks.
 A: This question make more sense when we don't limit outselves to real numbers.
For any $n = d + 1 > 1$ and $u = (u_0, u_1, \ldots, u_d) \in \mathbb{C}^n$, consider following product
$$\Lambda(u_0,\ldots,u_d) = \prod_{(\epsilon_1,\ldots,\epsilon_d) \in \{ \pm 1 \}^d}
\left( \sqrt{u_0} + \sum_{k=1}^d \epsilon_k \sqrt{u_k} \right)\tag{*1}
$$
We can expand $\Lambda(\cdots)$ to a homogeneous polynomial in $\sqrt{u_k}$ of degree $2^d$
$$\Lambda(u_1,\ldots,u_d) = \sum_{(e_0,\ldots,e_d)\in \mathbb{N}^n}
A_{e_0,\ldots,e_d} \prod_{k=0}^d\sqrt{u_k}^{e_k} \tag{*2}
$$
whose coefficients $A_{e_0,\ldots,e_d} \in \mathbb{Z}$ and vanish unless $e_0 + \ldots + e_d = 2^d$.
Consider the effect of flipping the sign of $\epsilon_\ell$ for some $\ell \ge 1$ in $\Lambda(\cdots)$.  


*

*In $(*1)$, this rearrange the order of product but leaves the value of $\Lambda(\ldots)$ untouched.   

*In $(*2)$, the coefficient $A_{e_0,\ldots,e_d}$ picks up a factor $(-1)^{e_\ell}$. 


Since the value of product doesn't change, $A_{e_0,\ldots,e_d}$ vanishes unless $e_\ell$ is even. Since this is true for every $\ell \ge 1$ and $A_{e_0,\ldots,e_d}$ vanishes unless $e_0 + \cdots + e_d = 2^d$, $A_{e_0,\ldots,e_d}$ also vanishes unless $e_0$ is even. This implies in expansion
$(*2)$, all square roots get completed. 
As a result, $\Lambda(\cdots)$ is a homogeneous polynomial in $u_0,\ldots, u_d$ of degree $2^{d-1}$:
$$\Lambda(u_1,\ldots,u_d) = \sum_{(e_0,\ldots,e_d)\in \mathbb{N}^n}
B_{e_0,\ldots,e_d} \prod_{k=0}^du_k^{e_k} \tag{*3}
$$
whose coefficients $B_{e_0,\ldots,e_d} \in \mathbb{Z}$ and vanish unless $e_0 + \ldots + e_d = 2^{d-1}$.
If $\sqrt{u_0} \pm \sqrt{u_1} \pm \cdots \pm \sqrt{u_d} = 0$ for any choice of sign of the square roots, then by construction, $u_0, \ldots, u_d$ need to satisfy the polynomial equation $\Lambda(u_0,\ldots,u_d) = 0$.
For the problem at hand, take $n = 3$ and substitute $(u_0,u_1,u_2)$ by $ (ax + \alpha, bx+\beta, cx+\gamma)$.
The equation $\sqrt{a x + \alpha} \pm \sqrt{b x + \beta} \pm \sqrt{cx + \gamma} = 0$ leads to a homogeneous polynomial equation in $ax, bx, cx, \alpha, \beta, \gamma$ of degree $2$:
$$\Lambda(ax+\alpha, bx+\beta, cx+\gamma) = 0$$
Expand this polynomial out against $x$, we obtain a quadratic equation in $x$:
$$C(\cdots) x^2 + D(\cdots) x + E(\cdots) = 0$$
It is easy to see the coefficients $C(\cdots)$ only depends on $(a,b,c)$. By setting $x$ to $1$ and $\alpha, \beta, \gamma$ to $0$, we find
$C(\cdots) = \Lambda(a,b,c)$. 
By setting $x$ to $0$, we find $E(\cdots) = \Lambda(\alpha,\beta,\gamma)$. This
leads to a equation of the form:
$$\Lambda(a,b,c)x^2 + D(\cdots)x + \Lambda(\alpha,\beta,\gamma) = 0$$
Now it comes to the mysterious condition $\sqrt{a} \pm \sqrt{b} \pm \sqrt{c} = 0$.
When this condition is fulfilled, $\Lambda(a,b,c) = 0$. Above equation simplifies
to a linear equation in $x$.
$$D(\cdots)x + \Lambda(\alpha,\beta,\gamma) = 0$$
We can determine the last unknown coefficient $D(\cdots)$ by setting $x$ to $1$.
At the end, we have

When  $\sqrt{a} \pm \sqrt{b} \pm \sqrt{c} = 0$, then the equation
  $\sqrt{ax+\alpha} \pm \sqrt{bx+\beta} \pm \sqrt{cx+\gamma} = 0$ leads to
  a linear equation in $x$:
  $$\Lambda(a+\alpha,b+\beta,c+\gamma)x + \Lambda(\alpha,\beta,\gamma)(1-x) = 0$$
  where
  $$\Lambda(u,v,w) = u^2 + v^2 + w^2 - 2(uv+vw+uw)$$

In certain sense, one can argue this equation is simple because it's dependence on $x$ is linear. Unlike the general case where $\Lambda(a,b,c) \ne 0$, the solution for $x$ no longer involves any radicals.
Whether one agree this is simple is up to one's own judgement. To be honest, I don't.
