What is 1/((a-b)(a-c))+1/((b-c)(b-a))+1/((c-a)(c-b)) ?

The solution to 1/((a-b)(a-c))+1/((b-c)(b-a))+1/((c-a)(c-b)) is 0

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

simplify

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

0

Solution steps

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

Least Common Multiplier of

(a - b) (a - c), (b - c) (b - a), (c - a) (c - b) : - (a - b) (a - c) (b - c)

Adjust Fractions based on the LCM

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

Apply the fraction rule:

\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}

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

Simplify

- b + c - c + a - a + b : 0

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

Apply rule

\frac{0}{a} = 0

a \neq = 0

= 0

a^{2} - 2 a + 1

5 \leq 1 - n

1^{\frac{5}{3}}

14 = \frac{8}{3} (x + 6)

4 (2 x - 3) + 4 = 8 x - 8

What is 1/((a-b)(a-c))+1/((b-c)(b-a))+1/((c-a)(c-b)) ?
The solution to 1/((a-b)(a-c))+1/((b-c)(b-a))+1/((c-a)(c-b)) is 0