Given:

$x^{2}-2 a x+\left(a^{2}-b^{2}\right)=0$

On comparing it with $A x^{2}+B x+C=0$, we get:

$A=1, B=-2 a$ and $C=\left(a^{2}-b^{2}\right)$

Discriminant $D$ is given by:

$\begin{array}{l}
D=B^{2}-4 A C \\
=(-2 a)^{2}-4 \times 1 \times\left(a^{2}-b^{2}\right) \\
=4 a^{2}-4 a^{2}+4 b^{2} \\
=4 b^{2}>0
\end{array}$

Hence, the roots of the equation are real.

Roots $\alpha$ and $\beta$ are given by:
$\begin{array}{l}
\alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-(-2 a)+\sqrt{4 b^{2}}}{2 \times 1}=\frac{2 a+2 b}{2}=\frac{2(a+b}{2}=(a+b) \\
\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-(-2 a)-\sqrt{4 b^{2}}}{2 \times 1}=\frac{2 a-2 b}{2}=\frac{2(a-b)}{2}=(a-b)
\end{array}$

Hence, the roots of the equation are $(a+b)$ and $(a-b)$.