- #1
Ted7
- 4
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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!.
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!.