Question

Find the value of $\frac{a^2-b^2-c^2}{(a-b)(a-c)}+\frac{b^2-c^2-a^2}{(b-c)(b-a)}+\frac{c^2-a^2-b^2}{(c-a)(c-b)}$.

Answer 1

$\frac{a^2-b^2-c^2}{(a-b)(a-c)}$$+$$\frac{b^2-c^2-a^2}{(b-c)(b-a)}$$+$ $\frac{c^2-a^2-b^2}{(c-a)(c-b)}$

$=$$\frac{a^2-b^2-c^2}{(a-b)(a-c)}$ $-$ $\frac{b^2-c^2-a^2}{(b-c)(a-b)}$ $+$ $\frac{c^2-a^2-b^2}{(c-a)(c-b)}$

$=$$\frac{(a^2-b^2-c^2)(b-c)-(b^2-c^2-a^2)(a-c)+(c^2-a^2-b^2)(a-b)}{(a-b)(a-c)(b-c)}$

Expanding numerator and denominator we get,

$=$ $\frac{a^{2}b-a^{2}c-b^3+b^{2}c-c^{2}b+c^3-a{b}^2+b^{2}c+a{c}^2-c^3+a^3-a^{2}c+a{c}^2-b{c}^2-a^3+a^{2}b+b^3-a{b}^2}{a^{2}b-a^{2}c-abc+a{c}^2-a{b}^2+abc+b^{2}c-b{c}^2}$

$=$ $\frac{2(a^{2}b-a^{2}c+a{c}^2-a{b}^2+b^{2}c-b{c}^2)}{a^{2}b-a^{2}c+a{c}^2-a{b}^2+b^{2}c-b{c}^2}$

$=2$