Help with Algebra

Question

How do I solve the following equation in terms of $p$ and $q$ if,

$$a^2 + b^2 - ab = p$$

And

$$2a^3 + 2b^3 -3ab(a+b) = - q$$

I tried lots of stuff, but I just can't see it, I end up with complicated equations. Can anyone help?

Answer 1

Well here is a start which will get you through in the end. Though whether this is the best way, I don't know.

Multiply the first equation by $a+b$ to obtain $$a^3+b^3=p(a+b)$$

Substitute this in the second to obtain $$(2p-3ab)(a+b)=-q$$

Square this $$(2p-3ab)^2(a+b)^2=q^2$$

Now notice that the first equation can be written $$(a+b)^2=p+3ab$$

So that $$(2p-3ab)^2(p+3ab)=q^2$$

This is a cubic in the unknown $ab$ (or $3ab$) and can be solved using standard methods. Once $ab$ is known $(a+b)^2=p+3ab$ will give $a+b$ and $a$ and $b$ will emerge from a quadratic. Then check which of the solutions solve the original equations.

There may be an easier way through.

Answer 2

Not very pleasant, for sure, but you can consider the first equation and solve it for $a$ to get the two possible roots $$a_1=\frac{1}{2} \left(b-\sqrt{4 p-3 b^2}\right)\qquad \text{and} \qquad a_2=\frac{1}{2} \left(\sqrt{4 p-3 b^2}+b\right)$$

Replace in the second equation to get $$3 b^2 \sqrt{4 p-3 b^2}-p \sqrt{4 p-3 b^2}+q=0$$ $$-3 b^2 \sqrt{4 p-3 b^2}+p \sqrt{4 p-3 b^2}+q=0$$ Let $x^2=4p-3b^2$ to end with cubic equations in $x$.