Find all real numbers $x,y$ such that $x^3-y^3=7(x-y)$ and $x^3+y^3=5(x+y)$.
Problem
Source: Slovenia National MO 2005 2nd Grade P1
Tags: algebra, system of equations
MathArt4
05.04.2021 06:38
We have that $x^2+xy+y^2=7$ and $x^2-xy+y^2=5$. Subtracting the two equations we get that $2xy=2 \implies xy=1$. We also get that $(x+y)^2=7+xy \implies x+y=\pm 2\sqrt{2}$ and that $(x-y)^2=5-xy \implies x-y=\pm 2$. Now we simply must bash through all of our cases:
$$xy=1 \text{ and } x-y=2 \implies \boxed{(1-\sqrt{2},-1-\sqrt{2}) \text{ and } (1+\sqrt{2},\sqrt2-1)}$$$$xy=1 \text{ and } x-y=-2 \implies \boxed{(-1-\sqrt{2},1-\sqrt{2}) \text{ and } (\sqrt{2}-1,1+\sqrt{2})}$$There are, however, more solutions, if $x-y=0$ or $x+y=0$, since the LHS will be $0$ as well. So, we can substitute $x$ for $-y$ or $x$ for $y$, which gives us the other solution pairs $\boxed{(-\sqrt{5},-\sqrt{5}) \text{ and } (\sqrt{5},\sqrt{5})}$ as well as $\boxed{(-\sqrt{7},-\sqrt{7}) \text{ and } (\sqrt{7},\sqrt{7})}$. Finally, if both $x+y=0$ and $x-y=0$ we have the final solution pair of $\boxed{(0,0)}$.
teomihai
05.04.2021 06:51
jasperE3 wrote: Find all real numbers $x,y$ such that $x^3-y^3=7(x-y)$ and $x^3+y^3=5(x+y)$. hint:cases $x=y $ (solution pairs $\boxed{(-\sqrt{5},-\sqrt{5}) \text{ and } (\sqrt{5},\sqrt{5})}$ and$\boxed{(0,0)}$. ) or $x\neq{y}$ (and now $x=-y$ or $x\neq{-y} $)
oneteen11
05.04.2021 07:13
I think this problem should be in High School Math instead of High School Olympiads
samrocksnature
05.04.2021 07:15
Apparently this is for second-grade students.
a.shyam.25
05.04.2021 14:49
Just use sum of cubes and simplify expression