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Explore Stack InternalI've come across a very curious Ramanujan identity $$\sqrt[6]{7 \sqrt[3]{20} - 19} = \sqrt[3]{\frac{5}{3}} - \sqrt[3]{\frac{2}{3}}$$
You could probably prove this by taking the 6th power of both sides. But how could he have come to this clean result without extra terms left over? I'm overwhelmed at trying to imagine how this would be done.
Edit: Also, canhow would you generalize identities involving a 6th root as two 3rd roots?
I've come across a very curious Ramanujan identity $$\sqrt[6]{7 \sqrt[3]{20} - 19} = \sqrt[3]{\frac{5}{3}} - \sqrt[3]{\frac{2}{3}}$$
You could probably prove this by taking the 6th power of both sides. But how could he have come to this clean result without extra terms left over? I'm overwhelmed at trying to imagine how this would be done.
Edit: Also, can you generalize identities involving a 6th root as two 3rd roots?
I've come across a very curious Ramanujan identity $$\sqrt[6]{7 \sqrt[3]{20} - 19} = \sqrt[3]{\frac{5}{3}} - \sqrt[3]{\frac{2}{3}}$$
You could probably prove this by taking the 6th power of both sides. But how could he have come to this clean result without extra terms left over? I'm overwhelmed at trying to imagine how this would be done.
Edit: Also, how would you generalize identities involving a 6th root as two 3rd roots?
I've come across a very curious Ramanujan identity $$\sqrt[6]{7 \sqrt[3]{20} - 19} = \sqrt[3]{\frac{5}{3}} - \sqrt[3]{\frac{2}{3}}$$
You could probably prove this by taking the 6th power of both sides. But how could he have come to this clean result without extra terms left over? I'm overwhelmed at trying to imagine how this would be done.
Edit: Also, can you generalize identities involving a 6th root as two 3rd roots?
I've come across a very curious Ramanujan identity $$\sqrt[6]{7 \sqrt[3]{20} - 19} = \sqrt[3]{\frac{5}{3}} - \sqrt[3]{\frac{2}{3}}$$
You could probably prove this by taking the 6th power of both sides. But how could he have come to this clean result without extra terms left over? I'm overwhelmed at trying to imagine how this would be done.
I've come across a very curious Ramanujan identity $$\sqrt[6]{7 \sqrt[3]{20} - 19} = \sqrt[3]{\frac{5}{3}} - \sqrt[3]{\frac{2}{3}}$$
You could probably prove this by taking the 6th power of both sides. But how could he have come to this clean result without extra terms left over? I'm overwhelmed at trying to imagine how this would be done.
Edit: Also, can you generalize identities involving a 6th root as two 3rd roots?
I've come across a very curious Ramanujan identity $$\sqrt[6]{7 \sqrt[3]{20} - 19} = \sqrt[3]{\frac{5}{3}} - \sqrt[3]{\frac{2}{3}}$$
You could probably prove this by taking the 6th power of both sides. But how could he have come to this clean result without extra terms left over? I'm overwhelmed at trying to imagine how this would be done.
