How would I go about solving for $x$ in $\frac{(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}}{\sqrt{x-a}+\sqrt{x-b}}=a-b$? The question
This is a homework question. Given the following, I am to solve for $x$ in terms of $a$ and $b$:
$$\frac{(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}}{\sqrt{x-a}+\sqrt{x-b}}=a-b;a>b.$$
My attempt
Although I see the pattern of multiple occurrences of $(x-a)$, $(x-b)$ I can't see any way to simplify the fraction further, so I go on to simplify the expression by multiplying by $\sqrt{x-a}+\sqrt{x-b}$:
$$\begin{align*}
(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}&=(a-b)(\sqrt{x-a}+\sqrt{x-b})\\
&=a\sqrt{x-a}+a\sqrt{x-b}-b\sqrt{x-a}-b\sqrt{x-b}
\end{align*}$$
Now I have the following:
$$(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}=a\sqrt{x-a}+a\sqrt{x-b}-b\sqrt{x-a}-b\sqrt{x-b}$$
Simplifying the RHS as I was out of ideas at that point:
$$x\sqrt{x-a}-a\sqrt{x-a}+x\sqrt{x-b}-b\sqrt{x-b}=a\sqrt{x-a}+a\sqrt{x-b}-b\sqrt{x-a}-b\sqrt{x-b}$$
I noticed that all one of the common factors $\sqrt{x-a},\sqrt{x-b}$ so I tried to isolate them and factor them out -- that is, all $\sqrt{x-b}$ terms on one side and $\sqrt{x-a}$ terms on the other.
$$\sqrt{x-b}(x-a)=\sqrt{x-a}(2a-b-x)$$
I tried to then square both sides, but that led to quite a mess.
$$(x-b)(x^2-2ax+a^2)=(x-a)(4a^2-4ab+2bx-4ax+b^2+x^2)$$
I'm afraid to even begin trying to simplifying this. I'm convinced I'm going about it in the wrong way.
The $a>b$ hint is interesting, but I have no clue what implication it may have here.
I think the $(x-a)\sqrt {x-a}$ patterns may mean something, perhaps I could do something with $a\sqrt a=\sqrt{a^3}$, but at this point it is probably a dead end.
I appreciate any help.
 A: Use the formula: $a^3+b^3=(a+b)(a^2-ab+b^2)$.
We get:
$$\frac{(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}}{\sqrt{x-a}+\sqrt{x-b}}=\\
\frac{(\sqrt{x-a}+\sqrt{x-b})((x-a)-\sqrt{(x-a)(x-b)}+(x-b))}{\sqrt{x-a}+\sqrt{x-b}}=\\
2x-a-b-\sqrt{(x-a)(x-b)}=a-b \Rightarrow \\
(x-a)(x-b)=(2x-2a)^2 \Rightarrow \\
3x^2+(b-7a)x+4a^2-ab=0 \Rightarrow \\
x=\frac{(7a-b)\pm \sqrt{(b-7a)^2-12(4a^2-ab)}}{6}=\\
\frac{7a-b\pm (a-b)}{6}=\\
\frac{4a-b}{3}; a.$$
A: Hint: Define $$u=\sqrt{x-a}\\w=\sqrt{x-b}$$therefore $${w^3+u^3\over u+w}=w^2-u^2$$which yields to $$2u^3=uw^2-u^2w$$one answer is $u=0$ or $x=a$ which is valid. The others can be found by solving $$2u^2=w^2-uw$$or $$u^2+uw=a-b$$by substituting we obtain $$x-a+\sqrt{(x-a)(x-b)}=a-b$$
A: Hint: Write your equation in the form
$$\sqrt{x-a}(x+b-2a)=\sqrt{x-b}(a-x)$$ and square it. We get in the case of $$a>b$$ $$x=a$$ or $$x=\frac{1}{3}(4a-b)$$
A: Another way is as follows:


*

*Set $\boxed{x = a + t(a-b)}$ for $t \geq 0$
$$\begin{eqnarray*} 
\frac{(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}}{\sqrt{x-a}+\sqrt{x-b}} & = & a-b \\
& \Leftrightarrow & \\
\frac{t(a-b)\sqrt{t(a-b)}+(t+1)(a-b)\sqrt{(t+1)(a-b)}}{\sqrt{t(a-b)}+\sqrt{(t+1)(a-b)}} & = & a-b \\
& \Leftrightarrow & \\
\frac{t\sqrt{t}+(t+1)\sqrt{t+1}}{\sqrt{t}+\sqrt{t+1}} & = & 1 \\
& \Leftrightarrow & \\
(t(\sqrt{t+1} + \sqrt{t})+\sqrt{t+1})(\sqrt{t+1}-\sqrt{t}) & = & 1 \\
& \Leftrightarrow & \\
t+ t+1 - \sqrt{t(t+1)} & = & 1 \\
& \Leftrightarrow & \\
2t & = & \sqrt{t(t+1)} \\
& \Leftrightarrow & \\
t =\frac{1}{3} & \mbox{ or } & t= 0   \\
& \stackrel{x = a + t(a-b)}{\Leftrightarrow} & \\
\boxed{x = a + \frac{1}{3}(a-b)} &\mbox{ or } & \boxed{x= a} 
 \end{eqnarray*}$$
