Special case Part 1

Note that this is only possible for complex $b$ and $c$ , as the given conditions imply that $1=(b+c)^2 = b^2 +2bc+c^2 = b^2 + c^2 + 2$ , so $b^2+c^2 = -1$ .

Taking the left hand side of the equation we are to verify, $\frac{a}{b} + \frac{a}{c} = \frac{ac + ab}{bc} = \frac{a(b+c)}{1} = a(b+c)$ But, since $(b+c)^2 = 1$ , we know that $(b+c)=\pm1$ and, therefore, $a(b+c) = \frac{a}{b+c}$ . So the statement is True .