<s>[INST] Error in Polynomial Factoring 
From Ramanujan's Notebooks IV:
Let $\alpha,\beta$ and $\gamma$ be the roots of$$x^3-ax^2+bx-1=0\tag1$$Now, choose cube roots such that $(\alpha\beta\gamma)^{1/3}=1$ and then let$$z^3-\theta z^2+\varphi z-1=0\tag2$$Denote the cubic polynomial with roots $\alpha^{1/3},\beta^{1/3},\gamma^{1/3}$. Cubing both sides of the equality$$z^3-1=\theta z^2-\varphi z\tag3$$We find that$$(z^3-1)^3-\theta^3z^6+\varphi^3z^3+3\theta\varphi z^3(z^3-1)=0\tag4$$Since $\alpha^{1/3},\beta^{1/3},\gamma^{1/3}$ are roots of $(2)$, they are also the root of $(4)$. As a cubic polynomial in $z^3$, $(4)$ thus has the roots $\alpha,\beta,\gamma$. Comparing $(1)$ with $(4)$, we deduce$$a=\theta^3+3-3\theta\varphi\tag5$$$$b=\varphi^3+3-3\theta\varphi\tag6$$ If we define $t$ by$$\theta^3=a+6+3t\tag7$$Then, by $(2)$ and $(7)$, $$\alpha^{1/3}+\beta^{1/3}+\gamma^{1/3}=\theta=(a+6+3t)^{1/3}$$And similarly, $$(\alpha\beta)^{1/3}+(\beta\gamma)^{1/3}+(\gamma\alpha)^{1/3}=\varphi=(b+6+3t)^{1/3}$$

While the proof is fairly elementary, there is one problem that I'm not sure how Bruce got there. From $z^3-1=\theta z^2-\varphi z$, to the next step, how'd they get the $3\theta\varphi z^3(z^3-1)$. When you expand $\theta z^2-\varphi z$, you get$$(\theta z^2-\varphi z)^3=\theta^3z^6-3\theta^2\varphi z^5+3\theta\varphi^2z^4-\varphi^3z^3$$So where did that term come from?
 [/INST] The last term on the left multiplies out to $3\theta\varphi(z^3-1)=3\theta\varphi z^3-3\theta\varphi$ which are intended to be the two middle terms from cubing the original right side.  It should be $-3\theta\varphi z^3(\theta z^2-\varphi z)$
</s>