An equation relating sine of angles in a triangle I wanted to solve the following problem.

In $\triangle ABC$ we have $$\sin^2 A + \sin^2 B = \sin^2 C + \sin A \sin B \sin C.$$ Compute $\sin C$.

Since it's an equation for a triangle, I assumed that $\pi = A + B + C$ would be important to consider.
I've tried solving for $\sin C$ as a quadratic, rewriting $\sin C = \sin (\pi - A - B)$, but nothing seemed to work.
How does one approach this problem? Any help would be appreciated.
The answer is

$$\frac{2\sqrt{5}}{5}$$

 A: using the Theorem of sines we get
$$\sin^2(C)\left(\frac{a^2+b^2}{c^2}\right)=\sin^2(C)+\frac{ab}{c^2}\sin^3(C)$$
since $$\sin(C)\neq 0$$ we obtain
$$\sin(C)=\frac{a^2+b^2-c^2}{ab}$$ using $$2\cos(C)=\frac{a^2+b^2-c^2}{ab}$$ we get $$\tan(C)=2$$
A: By the law of sines the triangle $ABC$ is a similar to the triangle with sides-lengths $\sin{A}$, $\sin{B}$ and $\sin{C}$. 
Thus, by the law of cosines 
$$\sin^2C=\sin^2A+\sin^2B-2\sin{A}\sin{B}\cos{C}$$ and by the given
$$\sin^2C=\sin^2A+\sin^2B-\sin{A}\sin{B}\sin{C}.$$
Id est, $$2\cos{C}=\sin{C}$$ or
$$4(1-\sin^2C)=\sin^2C$$ or
$$\sin{C}=\frac{2}{\sqrt5}$$ or
$$C=\arcsin\frac{2}{\sqrt{5}}.$$
Done!
A: Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,
$$\sin^2A+\sin(B-C)\sin(B+C)=\sin A\sin B\sin C$$
As $\sin(B+C)=\sin(\pi-A)=\sin A$ and $\sin A>0$ cancelling it in both sides
$$\sin A+\sin(B-C)=\sin B\sin C$$
$$\sin(B+C)+\sin(B-C)=\sin B\sin C$$
$$\iff2\sin B\cos C=\sin B\sin C$$
As $\sin B>0,$  $$\dfrac{\sin C}2=\dfrac{\cos C}1=\pm\sqrt{\dfrac{\sin^2C+\cos^2C}{2^2+1^2}}=?$$
But again $\sin C>0$
