Question

Simplify the following algebraic expressions.

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

Answer 1

$$\frac{b+c}{(a-b)(a-c)}+\frac{c+a}{(b-a)(b-c)}+\frac{a+b}{(c-a)(c-b)}$$
Multiplying and dividing all three terms with ( $-1)$.
Also, multiplying and dividing first term with $(b-c)$, second term with (c - a), third term with $(a-b)$.
$$=\frac{(-1)(b-c)(b+c)}{(-1)(a-b)(a-c)}+\frac{(-1)(c-a)(c+a)}{(-1)(b-a)(b-c)}+\frac{(-1)(a-b)(a+b)}{(-1)(c-a)(c-b)}$$
$$=\frac{(c-b)(c+b)}{(a-b)(b-c)(c-a)}+\frac{(a-c)(a+c)}{(a-b)(b-c)(c-a)}+\frac{(b-a)(b+a)}{(a-b)(b-c)(c-a)}$$
$$=\frac{c^2-b^2}{(a-b)(b-c)(c-a)}+\frac{a^2-c^2}{(a-b)(b-c)(c-a)}+\frac{b^2-a^2}{(a-b)(b-c)(c-a)}$$
$$=\frac{(c^2-b^2)+(a^2-c^2)+(b^2-a^2)}{(a-b)(b-c)(c-a)}$$
$=0$
$$\therefore \frac{b+c}{(a-b)(a-c)}+\frac{c+a}{(b-a)(b-c)}+\frac{a+b}{(c-a)(c-b)}=0$$