Question

1/bc(b-a)(c-a)+1/ca(c-b)(a-b)+1/ab(a-c)(b-c)​

Answer 1

Answer:

(a−b)(a−c)

1

+

(b−c)(b−a)

1

+

(c−a)(c−b)

1

=0

Solution:

Given,

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

(a−b)(a−c)

1

+

(b−c)(b−a)

1

+

(c−a)(c−b)

1

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)⇒

(a−b)(a−c)

1

+(

(b−c)(−(a−b))

1

)+(

(−(a−c))(−(b−c))

1

)

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

(a−b)(a−c)

1

−

(b−c)(a−b)

1

+

(a−c)(b−c)

1

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

(a−b)(a−c)(b−c)

(b−c)−(a−c)+(a−b)

Grouping the terms,

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

(a−b)(a−c)(b−c)

b−c−a+c+a−b

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

(a−b)(a−c)(b−c)

0

=0

(0 divided by any number is 0)