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$\large \begin{cases} a + b + c = -abc \\ \dfrac 1a + \dfrac 1b + \dfrac 1c = 1 \end{cases}$

Given the above, where $a, b, c \ne 0$, find $\dfrac 1{a^2} + \dfrac 1{b^2} + \dfrac 1{c^2}$.

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