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

In summary, the system of equations can be simplified to (a-c)^2+(a^2-c^2)^2 = 160. This is an unsolved problem on the AOPS forum and Opalg provided a great insight to approach it.
  • #1
anemone
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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 \(\displaystyle (a-c)^2+(a^2-c^2)^2\).
 
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  • #2
anemone said:
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]
 
  • #3
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)
 
  • #4
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
 

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

What is the System of Equation Challenge (a+b)(b+c)=-1?

The System of Equation Challenge (a+b)(b+c)=-1 is a mathematical problem that involves solving for the values of a and b using the given equation.

How do I solve the System of Equation Challenge (a+b)(b+c)=-1?

To solve the equation, you can use the distributive property to expand the left side of the equation. Then, you can combine like terms and use algebraic techniques such as factoring or substitution to solve for the variables.

What are some common strategies for solving the System of Equation Challenge (a+b)(b+c)=-1?

Some common strategies for solving this equation include using the distributive property, isolating variables, and substituting one variable for another to create a simpler equation.

Is there a specific method I should use to solve the System of Equation Challenge (a+b)(b+c)=-1?

There is no one specific method that must be used to solve this equation. However, it is recommended to use a combination of algebraic techniques and to check your solution by plugging it back into the original equation.

What is the significance of the System of Equation Challenge (a+b)(b+c)=-1?

The System of Equation Challenge (a+b)(b+c)=-1 is a common example of a system of equations, which is a set of equations with multiple variables that need to be solved simultaneously. It is often used in mathematics to model real-world situations and to find the values of multiple unknown variables.

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