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?

The equation (4) is correct. After both sides of equation (3) are cubed and the terms on the right get moved to the left there is only zero on the right. Now the cube of the right side of (4) is expanded by binomial theorem to four terms. Two of them are combined and factored with one factor being the right side of (3) which is immediately replaced with the left side of (3). A simple example will make this clear. $$a = b-c,$$ $$a^3 = b^3 - 3b^2c +3bc^2 -c^3,$$ $$a^3 -b^3 +c^3 = -3bc(b - c),$$ $$a^3 - b^3 + c^3 = -3abc.$$ The author just combined several steps. Notice that if we change the sign of $b$ we get the pretty result that if $a+b+c=0$ then $a^3+b^3+c^3=3abc$, which is a fact that Ramanujan was very familiar with.
