Solve the following quadratic equations by factorization:

We have been given

$a^{2} b^{2} x^{2}+b^{2} x-a^{2} x-1=0$

$b^{2} x\left(a^{2} x+1\right)-1\left(a^{2} x+1\right)=0$

$\left(b^{2} x-1\right)\left(a^{2} x+1\right)=0$

Therefore,

$b^{2} x-1=0$

$b^{2} x=1$

$x=\frac{1}{b^{2}}$

or,

$a^{2} x+1=0$

$a^{2} x=-1$

$x=-\frac{1}{a^{2}}$

Hence, $x=\frac{1}{b^{2}}$ or $x=-\frac{1}{a^{2}}$.