Question

f(x,y) = x^3 - 3xy^2 and g(x,y) = 3x^2y - y^3. Find all solution (x,y) so that :
[f(x,y)]^2 - [g(x,y)]^2 = [f(x,y)]*[g(x,y)] = sqrt(2)

Answer 1

i hope this will help u
\[f^2-g^2=fg=\sqrt{2}\\ (f^2-g^2)^2=f^4+g^4-2f^2g^2\\f^4+g^4=(f^2-g^2)^2+2f^2g^2=2+2*2=6\\(f^2+g^2)^2=f^4+g^4+2f^2g^2=6+2*2=10\\f^2+g^2=\sqrt{10}\\(f+g)^2=f^2+g^2+2fg=\sqrt{10}+\sqrt{8}\\f+g=\pm(\sqrt{10}+\sqrt{8})\]

Answer 2

\[f-g=\frac{\sqrt{2}}{f+g}=\pm \frac{\sqrt{2}}{\sqrt{10}+\sqrt{8}}=\pm(\sqrt{5}-2)\]

Answer 3

\[f-g=x^3-3x^2y+3xy^2-y^3=(x-y)^3=\pm(\sqrt{5}-2)\\x=y \pm \sqrt[3]{\sqrt{5}-2}\]

Answer 4

\[2g=f+g-(f-g)=\pm(\sqrt{10}+\sqrt{8}-\sqrt{5}+2) \\ g=3x^2y-y^3=\pm \frac{1}{2}(\sqrt{10}+\sqrt{8}-\sqrt{5}+2) \\ 3(y \pm \sqrt[3]{\sqrt{5}-2})^2 y-y^3=\pm \frac{1}{2}(\sqrt{10}+\sqrt{8}-\sqrt{5}+2)\]

Answer 5

last equation represent 2 polynomial of degree 3 and we can solve it by computer and to be perfectly honest with u i dont like this method i will try to find a better one

Answer 6

oke, thank you... i understand, i will try to find the last answer

Answer 7

where this problem comes from? i mean is this a hw or something?

Answer 8

i got this problem from a group fb, its name ''soul-mate-matika jilid 1", i think this is a math olympiad problem, maybe...