[Algebra]
Problem: Consider the function $f(x)=x^{2}-2 x$. Determine all real numbers $x$ that satisfy the equation $f(f(f(x)))=3$.

Solution: Let $a=f(f(x))$.

Thus, the equation $f(f(f(x)))=3$ is equivalent to $f(a)=3$.

Since $f(a)=a^{2}-2 a$, then we obtain the equation $a^{2}-2 a=3$ which gives $a^{2}-2 a-3=0$ and $(a-3)(a+1)=0$.

Thus, $a=3$ or $a=-1$ which means that $f(f(x))=3$ or $f(f(x))=-1$.

Let $b=f(x)$.

Thus, the equations $f(f(x))=3$ and $f(f(x))=-1$ become $f(b)=3$ and $f(b)=-1$.

If $f(b)=3$, then $b=f(x)=3$ or $b=f(x)=-1$ using similar reasoning to above when $f(a)=3$.

If $f(b)=-1$, then $b^{2}-2 b=-1$ and so $b^{2}-2 b+1=0$ or $(b-1)^{2}=0$ which means that $b=f(x)=1$.

Thus, $f(x)=3$ or $f(x)=-1$ or $f(x)=1$.

If $f(x)=3$, then $x=3$ or $x=-1$ as above.

If $f(x)=-1$, then $x=1$ as above.

If $f(x)=1$, then $x^{2}-2 x=1$ and so $x^{2}-2 x-1=0$.

By the quadratic formula,

$$
x=\frac{-(-2) \pm \sqrt{(-2)^{2}-4(1)(-1)}}{2(1)}=\frac{2 \pm \sqrt{8}}{2}=1 \pm \sqrt{2}
$$

Therefore, the solutions to the equation $f(f(f(x)))=3$ are $x=3,1,-1,1+\sqrt{2}, 1-\sqrt{2}$.

Final answer: $3,1,-1,1+\sqrt{2}, 1-\sqrt{2}$