MHB System of Equation Challenge (a+b)(b+c)=-1

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The discussion revolves around solving a system of equations involving real numbers a, b, and c, specifically focusing on the expression (a-c)² + (a²-c²)². The equations provided include (a+b)(b+c) = -1, (a-b)² + (a²-b²)² = 85, and (b-c)² + (b²-c²)² = 75. Through algebraic manipulation, it is concluded that (a-c)² + (a²-c²)² equals 160. The thread highlights the complexity of the problem and expresses appreciation for the insights shared by a participant named Opalg. The solution is confirmed as correct and is noted to be an unsolved problem on the AOPS forum.
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Given that $$a,\,b$$ and $$c$$ are real numbers that satisfy the system of equations below:

$$(a+b)(b+c)=-1\\(a-b)^2+(a^2-b^2)^2=85\\(b-c)^2+(b^2-c^2)^2=75$$

Find $$(a-c)^2+(a^2-c^2)^2$$.
 
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anemone said:
Given that $$a,\,b$$ and $$c$$ are real numbers that satisfy the system of equations below:

$$(a+b)(b+c)=-1\\(a-b)^2+(a^2-b^2)^2=85\\(b-c)^2+(b^2-c^2)^2=75$$

Find $$(a-c)^2+(a^2-c^2)^2$$.
This took me much longer than it should have done!
[sp]$$\begin{aligned}(a-c)^2+(a^2-c^2)^2 &= \bigl((a-b) + (b-c)\bigr)^2 + \bigl((a^2-b^2) + (b^2-c^2)\bigr)^2 \\ &= (a-b)^2 + 2(a-b)(b-c) + (b-c)^2 + (a^2-b^2)^2 + 2(a^2-b^2)(b^2-c^2) + (b^2-c^2)^2 \\ &= (a-b)^2 + (a^2-b^2)^2 + (b-c)^2 + (b^2-c^2)^2 + 2(a-b)(b-c) + 2(a^2-b^2)(b^2-c^2) \\ &= 85 + 75 + 2(a-b)(b-c)\bigl(1 + (a+b)(b+c)\bigr) \\ &= 160 + 2(a-b)(b-c)(1 - 1) = 160 \end{aligned}$$

[/sp]
 
Awesome, as always, Opalg!(Yes)

This actually is an unsolved problem at AOPS forum, and I will share the link of this thread at their site now, and thanks again Opalg for your great insight to attack the problem!(Cool)
 
Opalg said:
This took me much longer than it should have done!
[sp]$$\begin{aligned}(a-c)^2+(a^2-c^2)^2 &= \bigl((a-b) + (b-c)\bigr)^2 + \bigl((a^2-b^2) + (b^2-c^2)\bigr)^2 \\ &= (a-b)^2 + 2(a-b)(b-c) + (b-c)^2 + (a^2-b^2)^2 + 2(a^2-b^2)(b^2-c^2) + (b^2-c^2)^2 \\ &= (a-b)^2 + (a^2-b^2)^2 + (b-c)^2 + (b^2-c^2)^2 + 2(a-b)(b-c) + 2(a^2-b^2)(b^2-c^2) \\ &= 85 + 75 + 2(a-b)(b-c)\bigl(1 + (a+b)(b+c)\bigr) \\ &= 160 + 2(a-b)(b-c)(1 - 1) = 160 \end{aligned}$$

[/sp]
Brilliant!

-Dan
 
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