7
$\begingroup$

Ramanujan found that

$$\begin{align*} & \sqrt[3]{(m^2+mn+n^2)\sqrt[3]{(m-n)(m+2n)(2m+n)}+3mn^2+n^3-m^3}\\ =&\sqrt[3]{\tfrac {(m-n)(m+2n)^2}9}-\sqrt[3]{\tfrac {(2m+n)(m-n)^2}9}+\sqrt[3]{\tfrac {(m+2n)(2m+n)^2}9} \end{align*}$$ for arbitrary $m$ and $n$. The problem is that I am not sure how to prove it.

Question: Is there a way to prove it?

Similarly to the Ramanujan's formula for $$\sqrt{\sqrt[3]{a}+\sqrt[3]{b}}$$ I tried starting with a polynomial and deriving above formula through some clever manipulation. Unfortunately, none of that worked. If I were to do it that way, the terms in the polynomial would have to be a square root.

The book provides its way of proof by cubing both sides brute-force and slugging out the difficult algebra. Is there another way to prove it?

$\endgroup$
3
  • $\begingroup$ Is there supposed to be an $a$ in the LHS of the equation? $\endgroup$ May 10, 2017 at 19:44
  • 1
    $\begingroup$ @vrugtehagel Sorry that was autocorrect automatically correcting $mn$ to "man" $\endgroup$
    – Crescendo
    May 10, 2017 at 20:05
  • 1
    $\begingroup$ I'd assume the right-hand side can be represented as the sum of the roots of a cubic and the left-hand side can be cleverly manipulated through polynomial substitutions and whatnot $\endgroup$
    – Frank W
    Jul 28, 2019 at 21:31

1 Answer 1

6
$\begingroup$

Denote $$c_1=\sqrt[3]{\tfrac {(m-n)(m+2n)^2}9},\>\>\> c_2=-\sqrt[3]{\tfrac {(2m+n)(m-n)^2}9},\>\>\> c_3=\sqrt[3]{\tfrac {(m+2n)(2m+n)^2}9} $$

and it is straightforward to verify that

\begin{align} c_1c_2c_3 &=- \frac{1}{9}(m-n)(2m+n)(m+2n)\\ c_1^3+c_2^3+c_3^3 &= \frac{1}{3}(m^3+6m^2n+3mn^2-n^3)\\ c_1^3c_2^3+c_2^3c_3^3+c_3^3c_1^3 &= \frac{1}{3} c_1c_2c_3(m^3-3m^2n-6mn^2-n^3)\\ \end{align} Next, let $$A=c_1+c_2+c_3, \>\>\> B=c_1c_2+c_2c_3+c_3c_1,\>\>\>C= c_1c_2c_3$$ and evaluate \begin{align} A^3 &= 3AB+ c_1^3+c_2^3+c_3^3-3c_1c_2c_3\\ &=3AB+ (m^3+3m^2n-n^3)\tag1 \\ B^3 &= 3c_1c_2c_3 AB + c_1^3c_2^3+c_2^3c_3^3+c_3^3c_1^3-3(c_1c_2c_3)^2\\ &=3(AB)C +(m^3-3mn^2-n^3)C\tag2 \end{align} and their product $$ A^3B^3 = 9(AB)^2C-27(AB)C^2+27C^3 -\frac13C(m^2+mn+n^2)^3\tag3 $$ where the followings are recognized in arriving at (3) \begin{align} & (m^3+3m^2n-n^3)+(m^3-3mn^2-n^3)=-9C \\ & (m^3+3m^2n-n^3)(m^3-3mn^2-n^3)=27C^3-\frac13(m^2+mn+n^2)^3\\ \end{align} Rearrange (3) $$(AB-3C)^3=-\frac13C(m^2+mn+n^2)^3$$ and substitute $AB$ via (1) to obtain the equation for $A$ $$[A^3-(3mn^2+n^3-m^3)]^3 = -9C(m^2+mn+n^2)^3$$ which leads to the Ramanujan formula $$A= \sqrt[3]{(m^2+mn+n^2)\sqrt[3]{(m-n)(m+2n)(2m+n)}+3mn^2+n^3-m^3} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.