Question

If \(x\) and \(y\) are real numbers so that \(x^{3} -3xy^{2}=44\) and \(y^{3} - 3x^{2}y=8\). find the value of \(x^{2} + y^{2}\)!

Answer 1

If I didn't make any mistakes I was able to get these relations so far:
(I'm not sure if they are helpful yet or not)
\[x^2+y^2=\frac{44x-8y}{(x-y)(x+y)} \text{ and also have } (x-y)(x^2+4xy+y^2)=44-8\]

Answer 2

there is another way I'm looking it also...
\[x^3-3xy^2=44 \rightarrow \frac{x^2}{y^2}=\frac{3x^3}{x^3-44} \\ y^3-3x^2y=8 \rightarrow \frac{x^2}{y^2} =\frac{y^3-8}{3y^3}\]

Answer 3

\[x^3-3xy^2=44 \rightarrow x^2-3y^2=\frac{44}{x} \\ y^3-3x^2y=8 \rightarrow y^2-3x^2=\frac{8}{y} \\ \text{ combining the two } \\ x^2+y^2-3(x^2+y^2)=\frac{8}{y}+\frac{44}{x} \\ -2(x^2+y^2)=\frac{8}{y}+\frac{44}{x} \\ x^2+y^2=\frac{-1}{2}(\frac{8}{y}+\frac{44}{x})\]

Answer 4

any ideas @chihiroasleaf from anything I have said or anything you have thought of ?

Answer 5

*

Answer 6

this, what I'e got so far

\[x^3-3xy^2=44 \rightarrow y^2=\frac{x^3-44}{3x}\]
\[y^3-3x^2y=8 \]
\[y(y^2-3x^2)=8\]
\[\sqrt{\frac{x^3-44}{3x}} \left( \frac{x^3-44}{3x}-3x^2\right)=8\]
\[\sqrt{\frac{x^3-44}{3x}} \left( \frac{x^3-44-9x^3}{3x}\right)=8\]
\[\sqrt{\frac{x^3-44}{3x}} \left( \frac{-44-8x^3}{3x}\right)=8\]

Then, I doubt to continue this step :D

is There might be easier step?

Answer 7

\[x^3-3xy^2=44\quad,\quad y^3-3x^2y=8\]\[(x+iy)^3=x^3+3x^2(iy)+3x(iy)^2+(iy)^3=x^3+i3x^2y-3xy^2-iy^3=\]\[=x^3-3xy^2-i(y^3-3x^2y)=44-8i\]\[|(x+iy)^3|=|44-8i|\]\[|x+iy|^3=|44-8i|\]\[(x^2+y^2)^{3/2}=|44-8i|\]\[x^2+y^2=|44-8i|^{2/3}=(44^2+8^2)^{1/3}=2000^{1/3}=\sqrt[3]{2000}\]

Answer 8

@nikvist could you explain the second last step?