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$\large{ \begin{cases} a+a^2+a^3+\cdots=\dfrac{5}{6} \\ a^2+a^3+a^4+\cdots=\dfrac{25}{66} \\ b+b^2+b^3+\cdots=\dfrac{6}{5} \\ b^2+b^3+b^4+\cdots=\dfrac{36}{55} \end{cases}}$

Given the above, find $a^2-b^2$.

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