If \[a + b + c = 0\] , prove that $\dfrac{{{a^2}}}{{bc}} + \dfrac{{{b^2}}}{{ca}} + \dfrac{{{c^2}}}{{ab}} = 3$

Hint:
We can rearrange the given equation and write each variable in terms of the other variables. Then we can take the LHS of the equation we need to prove and find its LCM. Then we can expand the numerator using the expansion of the cube of sum of three numbers. Then we can substitution and cancel the common terms. Then we will obtain the RHS and prove the statement.

Complete step by step solution:
We are given that \[a + b + c = 0\] …. (1)
We can rearrange (1) to write a in terms of other two,
 \[ \Rightarrow a = - \left( {b + c} \right)\] … (2)
Similarly, we can write b,
 \[ \Rightarrow b = - \left( {c + a} \right)\] …. (3)
Now we can write c as,
 \[ \Rightarrow c = - \left( {a + b} \right)\] … (4).
We need to prove that $\dfrac{{{a^2}}}{{bc}} + \dfrac{{{b^2}}}{{ca}} + \dfrac{{{c^2}}}{{ab}} = 3$ .
We can take the LHS.
 $ \Rightarrow LHS = \dfrac{{{a^2}}}{{bc}} + \dfrac{{{b^2}}}{{ca}} + \dfrac{{{c^2}}}{{ab}}$
We can find the LCM of the LHS. For that we can multiply the terms with a, b and c,
 $ \Rightarrow LHS = \dfrac{{{a^3}}}{{abc}} + \dfrac{{{b^3}}}{{abc}} + \dfrac{{{c^3}}}{{abc}}$
As the denominators are same, we can add the numerators,
 $ \Rightarrow LHS = \dfrac{{{a^3} + {b^3} + {c^3}}}{{abc}}$
We know that ${\left( {a + b + c} \right)^3} = {a^3} + {b^3} + {c^3} + 3\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)$ ,
On rearranging this, we get,
 ${a^3} + {b^3} + {c^3} = {\left( {a + b + c} \right)^3} - 3\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)$
On substituting this on the LHS, we get,
 $ \Rightarrow LHS = \dfrac{{{{\left( {a + b + c} \right)}^3} - 3\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)}}{{abc}}$
On substituting equations, (1), (2) and (3), we get,
 $ \Rightarrow LHS = \dfrac{{{{\left( 0 \right)}^3} - 3\left( { - c} \right)\left( { - a} \right)\left( { - b} \right)}}{{abc}}$
On simplification and cancelling the negative signs, we get,
 $ \Rightarrow LHS = \dfrac{{3abc}}{{abc}}$
On cancelling the common terms, we get,
 $ \Rightarrow LHS = 3$
But we have $RHS = 3$
 $ \Rightarrow LHS = RHS$
Hence, if \[a + b + c = 0\] , then $\dfrac{{{a^2}}}{{bc}} + \dfrac{{{b^2}}}{{ca}} + \dfrac{{{c^2}}}{{ab}} = 3$

So, the required equation is proved.

Note:
We must know the expansion ${\left( {a + b + c} \right)^3} = {a^3} + {b^3} + {c^3} + 3\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)$ . We cannot write the sum of the cubes as the cube of the sums. While writing one variable in terms of the other two variables in the 1st step, we must take care of the sign. While substituting this on the LHS also, we must substitute with the sign and cancel out the negative sign. We must understand that the negative and negative will give positive when multiplied.