A 3 − 3 A 2 + 5 A − 1 = B 3 − 3 B 2 + 5 B − 5 = 0 .
Given thatFind the real value of A + B .
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There's a simpler approach to this. Hint: show that the two cubic equations are related by a suitable substitution.
Could you explain this step by step?
See my solution to an equivalent problem: Cubic Twins
Clearly, the curve has a point symmetry about the inflection point (1,2) as
f(1+x) +f(1-x) = 4.
Hence, the sum of roots A +B =2.
Hm, I fail to understand what you are trying to say. What is f ( x ) ? Unfortunately, not all of us are mind-readers, and we can only read what you have written down.
Hm, I fail to understand what you are trying to say. What is f ( x ) ? Unfortunately, not all of us are mind-readers, and we can only read what you have written down.
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I have taken f(x) = x^3 -3x^2 +5x - 1
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With that as f ( x ) , I agree that we do have f ( 1 + x ) + f ( 1 − x ) = 4 by expanding.
How does that tell us that A + B = 2 ?
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Define f ( x ) = x 3 − 3 x 2 + 5 x = ( x − 1 ) 3 + 2 ( x − 1 ) + 3 .
f ( A ) − 3 = ( A − 1 ) 3 + 2 ( A − 1 )
f ( B ) − 3 = ( B − 1 ) 3 + 2 ( B − 1 )
Note that f ( A ) + f ( B ) − 6 = A 3 − 3 A 2 + 5 A + B 3 − 3 B 2 + 5 B − 6 = 0 ,
and f ( A ) + f ( B ) − 6 = ( A − 1 ) 3 + ( B − 1 ) 3 + 2 ( A + B − 2 ) .
So ( A + B − 2 ) [ ( A − 1 ) 2 − ( A − 1 ) ( B − 1 ) + ( B − 1 ) 2 + 2 ] = 0 .
Note that ( A − 1 ) 2 − ( A − 1 ) ( B − 1 ) + ( B − 1 ) 2 + 2 ≥ 2 .
Hence, A + B = 2 .