VerifiedHint: Let’s expand the equation and eliminate. By eliminating we can come to the answer.
Complete step by step answer:
First of all we make the denominator common so that it is easy to take the LCM
$\dfrac{{{a^2} - {b^2} - {c^2}}}{{(a - b)(a - c)}} - \dfrac{{{b^2} - {c^2} - {a^2}}}{{(b - c)(a - b)}} + \dfrac{{{c^2} - {a^2} - {b^2}}}{{(a - c)(b - c)}}$
Now we take LCM of denominator to solving the fraction,
$\dfrac{{(b - c)({a^2} - {b^2} - {c^2}) - (a - c)({b^2} - {c^2} - {a^2}) + (a - b)({c^2} - {a^2} - {b^2})}}{{(a - b)(a - c)(b - c)}}$
First we multiply the numerator and get the result.
${a^2}b - {b^3} - {c^2}b - {a^2}c + {b^2}c + {c^3} - a{b^2} + a{c^2} + {a^3} + {b^2}c - {c^3} - {a^2}c + a{c^2} - {a^3} - a{b^2} - b{c^2} + {a^2}b + {b^3}$
By solving this we get,
${\text{2}}{{\text{a}}^2}b - 2{c^2}b - 2{a^2}c + 2{b^2}c - 2a{b^2} + 2a{c^2}$
Now take 2 common out of the equation,
${\text{2(}}{{\text{a}}^2}b - {c^2}b - {a^2}c + {b^2}c - a{b^2} + a{c^2})$
Now let's multiply the denominator,
$(a - b)(ab - ac - bc + {c^2})$
$({a^2}b - {a^2}c - abc + a{c^2} - a{b^2} + abc + {b^2}c - b{c^2})$
By solving the denominator we get,
${\text{(}}{{\text{a}}^2}b - {c^2}b - {a^2}c + {b^2}c - a{b^2} + a{c^2})$
Now write the numerator and denominator at their place,
$\dfrac{{{\text{2(}}{{\text{a}}^2}b - {c^2}b - {a^2}c + {b^2}c - a{b^2} + a{c^2})}}{{{\text{(}}{{\text{a}}^2}b - {c^2}b - {a^2}c + {b^2}c - a{b^2} + a{c^2})}}$
By eliminating this we get 2 as answer,
$\therefore {\text{Answer = 2}}$
Note: - We have to take care of simplifications carefully in these kinds of problems. Since we are dealing with big equations, there are lots of chances of making mistakes.
