# Now, just square it

**Algebra**Level 3

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

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

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

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

×

Problem Loading...

Note Loading...

Set Loading...