Isn't 100000 such a huge number (Part-8)?

Expanding the brackets and transferring all the terms to one side, we get:

$a^{100000}+a^{2}+b^{100000}+b^{2}+c^{100000}+c^{2}-2a^{50000}b-2b^{50000}c-2c^{50000}a=0$

Rearranging the terms, we get:

$(a^{50000}-b)^{2}+(b^{50000}-c)^{2}+(c^{50000}-a)^{2}=0$ .

We can deduce from here that each and every term of the above equation is equal to zero as they are rational.

So, $a^{50000}=b, b^{50000}=c$ and $c^{50000}=a$ .

By substitution, we get:

$((a^{50000})^{50000})^{50000})=a$ .

$a^{125000000000000}=a$ .

Clearly, $a=0$ is a solution to this equation.

Now consider the case when $a≠0$ :

We can divide both sides by $a$ to get $a^{124999999999999}=1$ .

Now, we get $124999999999999$ complex roots of unity and $1$ solution when $a=0$ .

So, in all there are $125000000000000$ ordered triplets which can be represented as $12500\times 10^{10}$ .

Hence, our final answer is $\color{#D61F06}{\huge{12500.}}$