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$\begin{cases} a+a^2+a^3+\ldots = \dfrac{5}{6} &\boxed{1}\\ a^2+a^3+a^4+\ldots=\dfrac{25}{66} &\boxed{2}\\ b+b^2+b^3+\ldots=\dfrac{6}{5} &\boxed{3}\\ b^2+b^3+b^4+\ldots=\dfrac{36}{55} &\boxed{4}\end{cases}$

$\boxed{1} - \boxed{2}$ :

$\left(a+a^2+a^3+\ldots \right) - \left(a^2+a^3+a^4+\ldots\right) = \dfrac{5}{6} - \dfrac{25}{66}\\ a=\dfrac{30}{66} = \dfrac{5}{11}$

$\boxed{3} - \boxed{4}$ :

$\left(b+b^2+b^3+\ldots \right) - \left(b^2+b^3+b^4+\ldots\right) = \dfrac{6}{5} - \dfrac{36}{55}\\ b=\dfrac{30}{55} = \dfrac{6}{11}$

$a^2-b^2\\ =(a-b)(a+b)\\ =\left(\dfrac{5}{11}-\dfrac{6}{11}\right)\left(\dfrac{5}{11}+\dfrac{6}{11}\right)\\ =\left(-\dfrac{1}{11}\right)\left(\dfrac{11}{11}\right)\\ =\boxed{-\dfrac{1}{11}}$

Substitute the second equation by taking a^2 common and in the fourth equation b^2 common and then substitute the values of first and third equation in them.you will get a^2 as 5/11 and b^2 as 6/11. Subtracting them gives the answer -1/11

Multiply a in first equation and then compare with second equation to get $a=\frac {5}{11}$ and similarly multiply b to third to get $b=\frac {6}{11}$ DETAILED SOLUTION: $a^{2}+...=\frac {5a}{6} and a^{2}+...=\frac {25}{66}.$ Since lhs is same, $\frac {5a}{6}=\frac {25}{66} or a=\frac {5}{11}$ and similarly find b.