Comparing Two Numbers with Fractional Exponents

0:00 Introduction. Substitution method used in this video.
1:01 Problem (a).
4:08 Problem (b), of which the way of organizing the expressions is quite different from the previous problem.
The type of problems that was rarely featured in my channel: The comparison problem. This video also features quite lengthy algebra involving polynomials with maximum degree of 6. So it's definitely worth watching it, right?
#algebra #exponents #inequalities #comparison
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[@Limited_Light]

You could use (x^2 + xy + y^2)^2
= (x^2 + 2xy + y^2 - xy)^2
= ((x + y)^2 - xy)^2
= (x + y)^4 - 2xy(x + y)^2 + (xy)^2 ... .

[@CornerstonesOfMath]

I think it could've been better if you elaborated how your end result helps solving this particular problem, because I think it actually does help.
That is, for that lengthy (x - y)(x + y)^3 - (x^2 + xy + y^2)^2 part where I have expanded everything, using your result,
(x - y)(x + y)^3 - (x^2 + xy + y^2)^2
= (x - y)(x + y)^3 - (x + y)^4 + 2xy(x + y)^2 - (xy)^2
= (x - y - (x + y))(x + y)^3 + 2xy(x + y)^2 - (xy)^2
= - 2y(x + y)^3 + 2xy(x + y)^2 - (xy)^2
= (- 2y(x + y) + 2xy)(x + y)^2 - (xy)^2
= - 2(y^2)(x + y)^2 - (x^2)(y^2)
= - (y^2)(2(x + y)^2 + x^2)
= - (y^2)(3x^2 + 4xy + 2y^2),
so it actually leads to desired simplification without expanding the whole thing.

[@Limited_Light]

@@CornerstonesOfMath I didn't go further because I was tired (giving and grading final exams this week) and wasn't entirely sure that continuing would help. I just suspected it. In the past few months, for things I want to film & post to my educational cahnnel, I often ran into things like that.