Simultaneous equation involving cubics

Question

I have been trying to solve these simultaneous equations but have been unable to find success. I had to resort to using the online graphing tool known as desmos to find out the values of $x$ and $y$. I would really appreciate a step by step guide to solving these equations from anyone who is willing please, as no matter what I use: substitution, elimination or the difference of two squares I am unable to get an answer.

$$ \left\{ \begin{array}{cc} (x+y)(x^2-y^2) &= 4 \\ (x-y)(x^2 +y^2) &= \frac{5}{2} \end{array} \right. $$

Graph of the simultaneous equations

according to desmos the answers are: $(x,y)=(-0.5, -1.5)$ or $(1.5,0.5)$

Thank you to anyone who is willing to help with this deceptively simple-looking set of simultaneous equations.

Answer 1

$$(x+y)(x^2-y^2) = 4 \tag 1$$ $$(x-y)(x^2 +y^2)= \frac{5}{2}\tag 2$$

As @GerryMyerson suggested in comments, let $y=k x$. Simplify and divide $(1)$ by $(2)$ to obtain $$\frac{(k+1)^2}{k^2+1}=\frac{8}{5}\quad \implies \quad 3 k^2-10 k+3=0$$ giving as roots $k_1=\frac 13$ and $k_2=3$.

Use $y=3x$ and replace in $(1)$ to get $$-32x^3=4 \quad \implies x=-\frac 12 \quad\text{and}\quad y=-\frac 32 $$

Use $y=\frac x3$ and replace in $(1)$ to get $$\frac{32 }{27}x^3=4\quad \implies x=\frac 32 \quad\text{and}\quad y=\frac 12 $$

The four complex roots have not been considered.

Answer 2

We have $$(x+y)(x^2-y^2)=4\\(x-y)(x^2+y^2)=\frac52$$

so $$(x+y)(x+y)(x-y)=4\\2(x-y)(x^2+y^2)=5$$

which is equivalent to $$2(x-y)(x^2+y^2)=(x+y)^2(x-y)+1\Rightarrow\\(x-y)[2(x^2+y^2)-(x+y)^2]=1\Rightarrow\\(x-y)(x^2+y^2-2xy)=1\Rightarrow\\(x-y)(x-y)^2=1\Rightarrow\\(x-y)^3=1$$

If we are confining our solutions to the reals, we have $$x=1+y$$

By substituting in say, the second equation we obtain $$(1+y-y)[(1+y)^2+y^2]=\frac52\Rightarrow\\4y^2+4y-3=0$$

Now, by solving the quadratic, we find the desired pair of solutions $(x,y)=\large(\frac32,\frac12)$ or $(x,y)=\large(-\frac12,-\frac32)$

Answer 3

Let $u=x+y, v=x-y$, then $u^2v=4, v(u^2+v^2)/2=\frac52 \implies u^2v+v^3=5\implies v^3=1$.

If one is looking only for real solutions, $v=1, u=\pm2$, gives $(x, y)\in \{(\frac32, -\frac12), (-\frac12, \frac32)\}$.

For the complex ones, use $v=\omega, \omega^2$ where $\omega$ is a non-real cube root of unity.

Answer 4

Hint.

Making $x = \frac 12(u+v),\ \ y = \frac 12(u-v)$ we have

$$ \cases{ u^2 v = 4\\ u^2 v +v^3= 5 }\Rightarrow v^3=1,\ \ u = \frac 12,\ \ \text{etc.} $$