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

[anemone]

Given that \(\displaystyle a,\,b\) and \(\displaystyle c\) are real numbers that satisfy the system of equations below:

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

Find \(\displaystyle (a-c)^2+(a^2-c^2)^2\).

 

[Opalg]

Given that \(\displaystyle a,\,b\) and \(\displaystyle c\) are real numbers that satisfy the system of equations below:

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

Find \(\displaystyle (a-c)^2+(a^2-c^2)^2\).

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]

 

[anemone]

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)

 

[topsquark]

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