(3) Let $0<x<1, a$ and $b$ be positive constants, then the minimum value of $\frac{a^{2}}{x}+\frac{b^{2}}{1-x}$ is $\qquad$ .
(3) $(a+b)^{2}$ Hint:
$$
\begin{aligned}
\frac{a^{2}}{x}+\frac{b^{2}}{1-x} & =\left(\frac{a^{2}}{x}+\frac{b^{2}}{1-x}\right)[x+(1-x)] \\
& =a^{2}+b^{2}+\frac{a^{2}(1-x)}{x}+\frac{b^{2} x}{1-x} \\
& \geqslant a^{2}+b^{2}+2 \sqrt{\frac{a^{2}(1-x)}{x} \cdot \frac{b^{2} x}{1-x}} \\
& =(a+b)^{2} .
\end{aligned}
$$