Proof of Ramanujan's Problem 525 with A=5, B=4

[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.
Thank you for your help!.

 

[mfb]

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.

 

[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 don't 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

 

[Ted7]

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