Question

1/(a-b)(a-c)+1/(b-c)(b-a)+1/(c-a)(c-b)

Answer 1

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Answer 2

$\frac{1}{(a-b)(a-c)}+\frac{1}{(b-c)(b-a)}+\frac{1}{(c-a)(c-b)}=0$

Solution:

Given,

$\frac{1}{(a-b)(a-c)}+\frac{1}{(b-c)(b-a)}+\frac{1}{(c-a)(c-b)}$

Simplifying the equation we get,

$\Rightarrow \frac{1}{(a-b)(a-c)}+\left(\frac{1}{(b-c)(-(a-b))}\right)+\left(\frac{1}{(-(a-c))(-(b-c))}\right)$

$\Rightarrow \frac{1}{(a-b)(a-c)}-\frac{1}{(b-c)(a-b)}+\frac{1}{(a-c)(b-c)}$

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

Grouping the terms,

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

$\Rightarrow \frac{0}{(a-b)(a-c)(b-c)} = 0$

(0 divided by any number is 0)