Brilliant! I solved it very differently: First i set a=33 for convenience and so the equation became: A^3 = m^3+n^3+3amn. I substituted a = m+n+delta on both sides, so i got: (m+n+delta)^3 = m^3 + n^3 + 3mn(m+n+delta). Opening the brackets and subtracting both sides by the lhs and rhs of the original equation, all terms left had delta in them so delta=0 derives the set of m+n=33 solutions. Dividing by delta we're left with a quadratic equation whose determinant is -3(m-n)^2, which eliminates all solutions where m is not equal to n. Now substituting m=n in our original equation, we are left with: A^3=2n^3+3an. Now using the same trick: setting this time a=delta-n, simplifying we get: Delta^2*(delta-3n) = 0. So either delta is 0 which implies a=-n=-m which gives the last solution, or a=2n=2m which means m+n=a which is already covered in previous solutions. Overall, for a general non zero value of a: Either m=n=-a - one solution, or m+n=a, which implies all solutions (i, a-i) where min(a,0)
Graphing the given equation will give you a line m+n-33=0 and a point (-33, -33). This is a very nice problem!
Nice my friend haha. Thanks for sharing👍👍👍
Brilliant!
I solved it very differently:
First i set a=33 for convenience and so the equation became:
A^3 = m^3+n^3+3amn.
I substituted a = m+n+delta on both sides, so i got:
(m+n+delta)^3 = m^3 + n^3 + 3mn(m+n+delta).
Opening the brackets and subtracting both sides by the lhs and rhs of the original equation, all terms left had delta in them so delta=0 derives the set of m+n=33 solutions.
Dividing by delta we're left with a quadratic equation whose determinant is -3(m-n)^2, which eliminates all solutions where m is not equal to n.
Now substituting m=n in our original equation, we are left with:
A^3=2n^3+3an.
Now using the same trick: setting this time a=delta-n, simplifying we get:
Delta^2*(delta-3n) = 0.
So either delta is 0 which implies a=-n=-m which gives the last solution, or a=2n=2m which means m+n=a which is already covered in previous solutions.
Overall, for a general non zero value of a:
Either m=n=-a - one solution,
or m+n=a, which implies all solutions (i, a-i) where min(a,0)
Thats really nice my friend haha👍👍👍 Nice method!
you have a talent to explain complex sums so easily…gives me a lot of motiv:)
I agree with you
Wow thanks a lot for your compliment my friend👍👍👍
Or just use from the start one of euler's identity a^3+b^3+c^3-3abc= {(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]}/2
For a=m b=n c=-33
Isn't that vieta's formula?
@Min-cv7nt i learned it as a factoring method for a^3+b^3+c^3 my book said it was from euler but I couldn't find it on the internet
That is a most excellent idea! Thanks for pointing this out.
Yes right haha. Thanks for sharing my friend👍👍👍
Great work as always professor🎉
Thanks a lot my friend👍👍👍
This is such a nice video
Thanks a lot my friend haha👍👍👍
What a brilliant and nice video professor
Thank you so much my friend👍👍👍
What a brilliant video
Thank you so much my friend👍👍👍
You are the best professor
Thanks a lot my friend haha👍👍👍
Can you do ploynomial divison after you find the factor m+n-33 or no?
Yes, that's what I did.
I also did this too
Yes nice haha👍👍👍