Mistakes give rise to Problems

Because the triples are of positive integers, we can cancel out $a$ from the numerator, because $a\neq 0$ as $a$ is asked to be POSITIVE INTEGER.

Thus we get $\dfrac{1}{b+c} = \dfrac{1}{b} + \dfrac{1}{c}$ .

But because $b$ and $c$ are also positive integers, we can say that $b+c > b \implies \dfrac{1}{b+c} < \dfrac{1}{b}$ $and$ $b+c>c \implies \dfrac{1}{b+c} < \dfrac{1}{c}$

And this results in $\dfrac{1}{b+c} < \dfrac{1}{b} + \dfrac{1}{c}........... \forall b,c \in \mathbb{N}$

Thus the required statement is not true for any triples $(a,b,c)$ , hence the answer is $\boxed{0}$

Working the expression and you'll get that the only possible solution could be when $a(c^2 + bc + b^2) = 0$ But a can't be 0, because the triples may only be positive integers. Note that $c^2 + bc + b^2 = 0$ if both b and c ar equal to 0, that's impossible because c can't be replaced by 0 on the first expression

Thus, there are no triples of positive integers such that the statement is true

Since $a$ , $b$ & $c$ are all positive integers : $\dfrac{a}{b+c} = \dfrac{a}{b} + \dfrac{a}{c}$

By multiplying both sides by $\frac{1}{a}$ , we get : $\dfrac{1}{b+c} = \dfrac{1}{b} + \dfrac{1}{c}$ $\therefore \ \ \dfrac{b+c}{b} + \dfrac{b+c}{c} = 1 = \dfrac{b}{b} + \dfrac{c}{b} + \dfrac{b}{c} + \dfrac{c}{c} = \dfrac{c}{b} + \dfrac{b}{c} +2$ $\therefore \ \ \dfrac{c}{b} + \dfrac{b}{c} = -1$ $\therefore \ \ \dfrac{c^2}{bc} + \dfrac{b^2}{bc} = \dfrac{b^2 + c^2}{bc} = -1$ $\therefore \ \ b^2 + c^2 = -bc$

But as we see , as both $b$ and $c$ are positive integers , Hence $b^2 + c^2$ is a positive value and also $-bc$ is a negative value . Which leads us to a contradiction and an impossible argument .

And thus , there is no triple $(a, b, c)$ (given the above conditions) satisfying the equation . And the answer is $\boxed{0}$ .

${\frac { a }{ b+c } =\frac { a }{ b } +\frac { a }{ c } \\ \frac { 1 }{ b+c } =\frac { 1 }{ b } +\frac { 1 }{ c } \\ \frac { 1 }{ b+c } =\frac { b+c }{ bc } \\ bc={ b }^{ 2 }+2bc+{ c }^{ 2 }\\ { b }^{ 2 }+bc+{ c }^{ 2 }=0}$

The equation above cannot be true for any positive integer as square of any positive integer is always positive. Also since both $b$ and $c$ are positive, $bc$ will always be positive.

Hence no triplet satisfies the equation.

Note that if there was a triple (a,b,c) that worked, then we could replace a with any other positive integer in order to make another working triple. But this would mean the answer is infinitely many and that can't be inputted, so the answer must be 0 :D.

But I actually did it the way it was meant to be in order to make sure i was right.