Mistakes give rise to problems- 2

case 1 is $a\neq 0$ .

If $a\neq 0$ , we have to find the triples such that $\dfrac{1}{b+c} = \dfrac{1}{b} + \dfrac{1}{c}$ and in $0 \leq b,c$ , always $\dfrac{1}{b+c} < \dfrac{1}{b}$ because the case of $c=0$ makes $\dfrac{1}{c}$ not defined .

Hence this gives us that case 2 , i.e. $a=0$ to be the only case.

For $a=0$ , $b$ and $c$ can be any pair of integers from $1$ to $10$ , hence there are $10$ choices each for $b$ and $c$ giving $10\times 10 = 100$ distinct ordered pairs $(b,c)$ , and adding a $0$ to each of them in the start will give $(a,b,c)$ , hence there will be $\boxed{100}$ triples following the property.

Another method to prove that there is no solution when $a\neq0$ :

$\frac{1}{b+c}=\frac{1}{b}+\frac{1}{c}$ implies that $bc=(b+c)^2$ or $b^2+c^2=-bc$ . Moreover, $b^2+c^2=-bc\leq0$ . Hence, $b=c=0$ which contradicts the given equation.

$(1)~\implies~b^2+c^2=- b*c. ~Since~b\nless 0 ~and~c\nless 0,~~this~ is ~not~ possible.\\ \text{Hove ever if both the sides are 0, then only this is possible.}\\ \text{This could happen only if a=0. And then b and c take any value, except 0.}\\ \text{So a=0 only and both b, c, 1 to 10, gives, }\\ 1*10*10=\color{#D61F06}{100}~ ~ORDERED~Triples.$

We have 2 cases.

$\text{Case 1: } {a = 0}$ $\text{In this case the equation is true for any values of b and c excluding 0}$ $\text{Therefore we have 100 solutions for case 1.}$

$\text{Case 2: } {a \neq 0}$ ${\frac{a}{b + c} = {\frac{a}{b} + \frac{a}{c} } }$ $\Rightarrow {\frac{1}{b + c} = {\frac{1}{b} + \frac{1}{c} } }$ $\Rightarrow { {\left(b+c\right)}^{2} = {bc}}$ $\Rightarrow {{b}^{2} + bc + {c}^{2} = 0}$ $\text{Which has no solutions when } {b > 0, c > 0}$

Therefore the answer is $\boxed{100}$

Case 1 : if a = 0 Then the given equation is true for any value of b and c in the given range, therefore triplets formed are 1×10×10 = 100 (b,c≠0). Moreover if a≠0, then formed condition is (b+c)^2 = bc which is only true if b,c=0 but this these values are neglected since it would render the given equation indeterminable. Thus total number of triplets is 100.

You're wrong suggesting it's 90, not because (0,b,b) and (0,c,c) are different, but because in counting to 100 they are not counted twice. (0,1,1...10) (0,2,1...10) (0,3,1...10)... See? The only bracket in my answer that leads to (0,1,1) is the first bracket and it only appears once. So 1x10x10 is suitable for receiving the correct answer of 100. Its good to spot that (a=0,b=1,c=1) only appears once so the ten cases where b=c do not appear twice and needn't be removed :)

Just be careful to preserve the original equation. (0, b, c) for 1 to 10 of b and c gives 10 x 10 = 100.

1 = 1 + 1 + c/ b + b/ c is impossible to be true for all positive numbers.

Answer: 100

if there was written unordered pairs then what is the difference in both of its answers

did the same way as Qi Huan Tan. this question could also be done by logic . we know from the question that the property is wrong and is supported(dependent) on the the denominator and hence it should mean for any value of "b" and "c" it is not true unless the value of the fraction is irrespective the value of "b" and "c" under the defined domain of "b"and "c" in the question . and the value of the fraction becomes independent of the "b"and "c" if "a=0". and then it is just the combination of a 10x10 square. =100