# I Ramanujan problem

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1. Jan 6, 2017

### Ted7

Hi everyone.
This is my proof (?)of ramanujan's problem 525: http://www.imsc.res.in/~rao/ramanujan/collectedpapers/question/q525.htm (link to problem)

[![enter image description here][1]][1]

$$\sqrt{A^{1/3}-B^{1/3}}=\frac{(A*B/10)^{1/3}+(A \times B)^{1/3}-(A^2)^{1/3}}{3} \Leftrightarrow \\ 9 \times (A^{1/3}-B^{1/3})=[(A*B/10)^{1/3}+(A \times B)^{1/3}-(A^2)^{1/3}]^2$$
for A=5 and B=4. we arrive to the final result

$$R=R \qquad (R=9 \times (A^{1/3}-B^{1/3}))$$

Is this proof correct?
If it isn't am I getting closer to the right answer?

[1]: https://i.stack.imgur.com/AP8hC.jpg
If you've seen this posted elsewhere ,notice that I posted it.

2. Jan 6, 2017

### Staff: Mentor

You left out all steps apart from one. If you can show that the two sides are equal, that works, but in general they are not equal. As an example, try A=8, B=0.

3. Jan 6, 2017

### MAGNIBORO

How did you come to the conclusion of:
$$\sqrt{A^{1/3}-B^{1/3}}=\frac{(A*B/10)^{1/3}+(A \times B)^{1/3}-(A^2)^{1/3}}{3}$$
the ecuation dont work for
the case A=1 , B=1
and the case of A= 27, and B=28
so is incorrect.

is like tell
$$A=A^{2}$$
for A=1 and A=0, we arrive to the final result
$$A=A^{2}$$
but this is obviously wrong, you can not "create" ecuations for Particular cases, If an equation is right
Must be true for all values of A

4. Jan 6, 2017

### Ted7

Thanks I was aware of it .I am going try to find the right equation ;)