A special equation
Nonzero real numbers a , b , and c are such that
⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ a 2 + b + c = a 1 b 2 + c + a = b 1 c 2 + a + b = c 1
Find ( a − b ) ( b − c ) ( c − a ) .
The answer is 0.
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1 solution
Your solution is incomplete. You have to prove that equality of two of a , b and c assures that all of them are real. In fact, I'm curious to know the values of a , b and c that satisfy the set of equations.
I have made an edit. Thx for saying this.
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Sorry, can you please explain this part that I don't understand, "then t=a+b+c=1 from the equation"?
If a , b , c are distinct, let t = a + b + c . The first equation becomes: a 2 + t − a = a 1 a 3 − a 2 + t a − 1 = 0 same for b and c .
As a , b , c are distinct, so The roots of x 3 − x 2 + t x − 1 = 0 are a , b , c
Then t = a + b + c = 1 from the equation, which becomes x 3 − x 2 + x − 1 = 0 , solving: x 3 − x 2 + x − 1 = 0 x 2 ( x − 1 ) + ( x − 1 ) = 0 ( x − 1 ) ( x 2 + 1 ) = 0 x = 1 As there is only one real solution, it is a contradiction. Which means at least two of a , b , c are equal.
Then, ( a − b ) ( b − c ) ( c − a ) = 0 .
P.S. To prove there is solution, let the case when a = b = c . Then, a 2 + 2 a = a 1 a 3 + 2 a 2 − 1 = 0 ( a + 1 ) ( a 2 + a − 1 ) = 0 Therefore, a = b = c = − 1 is one of the solution.