Help with this please

[Tincan]

This is the equation, I am having trouble with working or starting. Can someone help get me started, or work it so I might better get a grasp of the situation.
Thank you. This is not a joke.

a^2x^2-a^6x^4y^2-a^4x^6b^2+a^4b^2x^4y^2-a^2b^4x^2y^4+a^2b^4y^6+b^6x^2y^4-b^6y^6=0

 

[stapel]

a^2x^2-a^6x^4y^2-a^4x^6b^2+a^4b^2x^4y^2-a^2b^4x^2y^4+a^2b^4y^6+b^6x^2y^4-b^6y^6=0

Are you given values for "a" or "b"? Are you supposed to solve for x in terms of y, or for y in terms of x? Or did the instructions say something else entirely?

What tools or methods were covered recently in class? What have you tried? How far have you gotten?

Please be complete. Thank you!

Eliz.

 

[Tincan]

No instruction was given , this is extra credit. Have not started on it yet because I do not know how to even get started. The only thing we were told was to begin by grouping and no instruction on what variable to solve for. Thank you for the help.

 

[soroban]

Hello, Tincan!

Are you sure there are no typos?

If I change the first term, there is a neat solution . . .

. . . . \(\displaystyle \overbrace{a^6x^6}-a^6x^4y^2-a^4x^6b^2+a^4b^2x^4y^2-a^2b^4x^2y^4+a^2b^4y^6+b^6x^2y^4-b^6y^6\;=\;0\)

\(\displaystyle \text{Factor in pairs: }\;a^6x^4(x^2-y^2) \;-\; a^4b^2x^4(x^2-y^2) \;-\; a^2b^4y^4(x^2-y^2) \;+\; b^6y^4(x^2-y^2) \;=\;0\)

\(\displaystyle \text{Factor: }\;(x^2-y^2)\,\bigg[a^6x^4 - a^4b^4x^4 - a^2b^4y^4 + b^6y^4\bigg]\;=\;0\)

\(\displaystyle \text{Factor in pairs: }\;(x^2-y^2)\,\bigg[a^4x^4(a^2-b^2) - b^4y^4(a^2-b^2)\bigg] \;=\;0\)

\(\displaystyle \text{Factor: }\;(a^2-b^2)(x^2-y^2)\left(a^4x^4-b^4y^4\right) \;=\;0\)

\(\displaystyle \text{Factor: }\;(a^2-b^2)(x^2-y^2)(a^2x^2-b^2y^2)(a^2x^2+b^2y^2) \;=\;0\)

\(\displaystyle \text{Factor: }\;(a^2-b^2)(x-y)(x+y)(ax-by)(ax+by)(a^2x^2+b^2y^2) \;=\;0\)


\(\displaystyle \text{The solutions are: }\:\boxed{y \:=\:\pm x}\quad\boxed{y\:=\:\pm\frac{a}{b}x}\)

. . \(\displaystyle \bigg[\text{Should I include: }\;a \:=\:\pm b\;?\bigg]\)