Solve the following quadratic equations by factorization:

We have been given

$(a+b)^{2} x^{2}-4 a b x-(a-b)^{2}=0$

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

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

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

Therefore,

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

$(a+b)^{2} x=-(a-b)^{2}$

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

or,

$x-1=0$

$x=1$

Hence, $x=-\left(\frac{a-b}{a+b}\right)^{2}$ or $x=1$.