The value of a sum

Algebra Level 2

For real numbers a , b , c a,b,c , where a ≠ b , a ≠ c , b ≠ c a\ne b, a\ne c, b\ne c , there holds a b − c + b c − a + c a − b = 0 \dfrac a{b-c}+\dfrac b{c-a}+\dfrac c{a-b}=0 . The value of the sum a ( b − c ) 2 + b ( c − a ) 2 + c ( a − b ) 2 \dfrac a{(b-c)^2}+\dfrac b{(c-a)^2}+\dfrac c{(a-b)^2} is


The answer is 0.

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1 solution

From the condition there follows − a b − c = b c − a + c a − b -\dfrac a{b - c} = \dfrac b{c - a} + \dfrac c{a - b} . So − a ( b − c ) 2 = a − b − c ( a − b ) ( c − a ) -\dfrac a{(b - c)^2} = \dfrac{a-b-c}{(a - b)(c-a)} . Similarly, − b c − a = a b − c + c a − b -\dfrac b{c - a} = \dfrac a{b - c} + \dfrac c{a - b} implies − b ( c − a ) 2 = − a + b − c ( a − b ) ( b − c ) -\dfrac b{(c - a)^2} = \dfrac{-a + b - c}{(a - b)(b - c)} . Also, from − c a − b = a b − c + b c − a -\dfrac c{a - b} = \dfrac a{b - c} + \dfrac b{c - a} , we have − c ( a − b ) 2 = − a − b + c ( c − a ) ( b − c ) -\dfrac c{(a - b)^2} = \dfrac{-a - b + c}{(c - a)(b - c)} . Summing the left-hand sides and the right-hand sides, we obtain − a ( b − c ) 2 − b ( c − a ) 2 − c ( a − b ) 2 = a − b − c ( a − b ) ( c − a ) + − a + b − c ( a − b ) ( b − c ) + − a − b + c ( c − a ) ( b − c ) -\dfrac a{(b - c)^2} -\dfrac b{(c - a)^2} -\dfrac c{(a - b)^2} = \dfrac{a-b-c}{(a - b)(c-a)} + \dfrac{-a + b - c}{(a - b)(b - c)} + \dfrac{-a - b + c}{(c - a)(b - c)} . However, a − b − c ( a − b ) ( c − a ) + − a + b − c ( a − b ) ( b − c ) + − a − b + c ( c − a ) ( b − c ) = ( b − c ) ( a − b − c ) + ( c − a ) ( − a + b − c ) + ( a − b ) ( − a − b + c ) ( a − b ) ( b − c ) ( c − a ) = 0 \dfrac{a-b-c}{(a - b)(c-a)} + \dfrac{-a + b - c}{(a - b)(b - c)} + \dfrac{-a - b + c}{(c - a)(b - c)}= \dfrac{(b-c)(a-b-c)+ (c - a)(-a + b - c) +(a - b)(-a - b + c)}{(a - b)(b - c)(c - a)}=0 .

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