solve: Possible bug?

savecancel

Hello, @curios_mind!

I think you found an interesting problem. My appreciation is that $a=-5/8$ is the correct answer (up to complex representation). Let me explain...

Observe the numerator of your expression $f$ can be written as $$\sqrt[3]{2}\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)=\sqrt[3]{2}\bigl(\cos(\pi/3)-i\sin(\pi/3)\bigr)$$ which is nothing more than one of the possible solutions of the equation $x^3=-2$, or alternatively, this is one of the possible values of $\sqrt[3]{-2}$ in the complex plain.

Now, consider the denominator for $a=-5/8$: $$\left(-2\left(\frac{-5}{8}\right)-1\right)^{1/3}=\sqrt[3]{\frac{1}{4}}=\sqrt[3]{\frac{1}{2^2}}$$

So, what $f$ is for $a=-5/8$ turns out to be $$\frac{\sqrt[3]{-2}}{\sqrt[3]{\frac{1}{2^2}}}=\sqrt[3]{-2}\sqrt[3]{2^2}=\sqrt[3]{-2^3}=-2$$ So the original equation is satisfied.

If you want to be more formal, $$\frac{\sqrt[3]{-2}}{\sqrt[3]{\frac{1}{2^2}}}=\sqrt[3]{-2}\sqrt[3]{4}=\sqrt[3]{2}\bigl(\cos(\pi/3)-i\sin(\pi/3)\bigr)\sqrt[3]{4}\bigl(\cos(\pi/3)+i\sin(\pi/3)\bigr)=\sqrt[3]{8}(\cos^2(\pi/3)+\sin^2(\pi/3))=\sqrt[3]{8}=2$$

Of course, as you can see, this depends on you choosing the correct representation of complex numbers like $\sqrt[3]{-4}$. If you choose wrongly, for example, $\sqrt[3]{-4}=\sqrt[3]{4}\bigl(\cos(\pi/3)-i\sin(\pi/3)\bigr)$, you won't get the answer, but other possible value for $f$ in the complex plain. That is what happened when you subs $a=-5/8$: Sage didn't choose the correct complex representations of some cube roots.

In summary, the problem is that you have a complex equation with ambiguous numbers like cube roots. Sage and Maxima did what they could, which is to give you a correct answer, up to complex representation. In this case, it is up to you to decide if this is what you wanted or not. If this is not the solution you expected, or you want an alternative solution, I recommend you to rewrite the cube roots explicitly as one of their possible complex number values.

I hope this helps!