Help looking for solutions to these two equations?

Question

I need help looking for solutions of the form $(a,b)$ to the equations

1. $a^3 -ab +b^3 =0$
2. $3a^2 +a=3b^2 +b$

Any help would be much appreciated, thanks.

I've tried putting the two equations together, then solving for one variable as a cubic, but the equations become very complicated and messy. A similar problem occurs when I solve for $a$ in terms of $b$ in equation (2) then sub that into equation (1).

Answer 1

From equation (2) you get that $(a-b)(3(a+b)+1)=0$. Then, either $a=b$ or $3(a+b)+1=0$.

In the first case we get, putting $b=a$ in equation (1), that $a^2(2a-1)=0$. So, either $a=b=0$, or $a=b=1/2$.

Assume now that $a\neq b$. Then $b=-1/3-a$. Putting this in equation (1) you get that

$$a^3-a(-1/3-a)+(-1/3-a)^3=0$$

this s equivalent to $-\frac{1}{27}=0$, which has no solutions.

Now you can take these possible solutions and check if they satisfy the original system of equations.

Answer 2

(1)-(2) gives $a=b$ or $3(a+b)=-1$.

If $a=b$ then the first equation gives $a=b=0$ or $a=b=\frac{1}{2}$.

(1) gives $3(a+b)(a^2-ab+b^2)=3ab$. Substituting $3(a+b)=-1$ gives $(a+b)^2=0$, so $a+b=0$. Contradicting $3(a+b)=-1$.

So the only solutions are $(a,b)=(0,0)$ or $(\frac{1}{2},\frac{1}{2})$