a little help pls

Let $a$, $b$, and $c$, be nonzero real numbers such that $a+b+c=0$.
Compute the value of
$a\left(\dfrac 1b + \dfrac1c\right) + b\left(\dfrac1a + \dfrac1c\right) + c\left(\dfrac1a+\dfrac1b\right)$.

$\begin{array}{|rcll|} \hline a+b+c &=& 0 \\ a+b &=& -c \qquad (1) \\ a+c &=& -b \qquad (2) \\ b+c &=& -a \qquad (3) \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && \mathbf{a\left(\dfrac 1b + \dfrac1c\right) + b\left(\dfrac1a + \dfrac1c\right) + c\left(\dfrac1a+\dfrac1b\right)} \\\\ &=& \dfrac ab + \dfrac ac + \dfrac ba + \dfrac bc + \dfrac ca+\dfrac cb \\\\ &=& \dfrac ba + \dfrac ca + \dfrac ab + \dfrac cb + \dfrac ac+\dfrac bc \\\\ &=& \dfrac {b+c}a + \dfrac {a+c}b + \dfrac {a+b}c \\\\ &=& \dfrac {-a}a + \dfrac {-b}b + \dfrac {-c}c \\\\ &=& -1-1-1 \\\\ &=& \mathbf{-3} \\ \hline \end{array}$

$a\left(\dfrac 1b + \dfrac1c\right) + b\left(\dfrac1a + \dfrac1c\right) + c\left(\dfrac1a+\dfrac1b\right)=-3$