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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?

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Crescendo
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2 Answers2

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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.

Somos
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  • Is there a fourth power form similar to $a^3+b^3+c^3=3abc$? – Crescendo May 13 '17 at 18:28
  • Thanks for that question. Indeed there is. You want to look at power sums. This is, $s(n) := a^n+b^n+c^n$ given that $s(1) = a+b+c = 0$. Now, $s(2) = 2(a^2+ab+b^2), s(4)=s(2)^2/2, s(5)=s(3)s(2)5/6$ and so on. – Somos May 13 '17 at 19:44
  • I meant that $a^4+b^4+c^4+d^4$ for fourth powers, $a^5+b^5+c^5+d^5+e^5$ for fifth powers, and so on... Is there such formula for that? Sorry for not being too specific... – Crescendo May 14 '17 at 19:47
  • Thanks for that further question. Has a similar answer and you can come up with your own variation. Let $s(n):=a^n+b^n+c^n+d^n$ given that $s(1)=s(2)=0$. Now $s(5)=0, s(6)=s(3)^2/3, s(7)=s(3)s(4)7/12, s(8)=s(4)^2/4,s(9)=s(3)^3/9$ and so on. – Somos May 14 '17 at 21:11
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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 $-3\theta\varphi z^3(\theta z^2-\varphi z)$

Ross Millikan
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  • That's the problem I'm wondering. The author used $3\theta\varphi(z^3-1)$. Is it an error? And if so, what would be the correct procedure? – Crescendo May 11 '17 at 23:16
  • It is just expanding the cube of $\theta z^2-\varphi z$. The $\theta^3z^6$ and$ \varphi^3 z^3$ got moved to the other side. The two middle terms of the cube are as I show. – Ross Millikan May 11 '17 at 23:20
  • Perhaps I'm not quite understanding you. The author intended the $3\theta\varphi(z^3-1)$. Even though whenever I expand it, it isn't the same as the original function – Crescendo May 11 '17 at 23:26
  • I was just doing $(a-b)^3=a^3-b^3+3ab(a-b)$. Maybe there is some other relation that justifies the substitution – Ross Millikan May 12 '17 at 00:40
  • So the book is wrong... – Crescendo May 12 '17 at 01:13