The following result discussed by Ramanujan is very famous: 3√3√2−1=3√19−3√29+3√49 and can be easily proved by cubing both sides and using x=3√2 for simplified typing.

Ramanujan established many such denesting of radicals such as √5√15+5√45=5√1+5√2+5√8=5√16125+5√8125+5√2125−5√11253√5√325−5√275=5√125+5√325−5√9254√3+24√53−24√5=4√5+14√5−16√73√20−19=3√53−3√236√43√23−53√13=3√49−3√29+3√19

8√1±√1−(−1+√52)24=−1+√524√5±1√2

with the last one found in

Ramanujan’s Notebooks, Vol 5, p. 300. Most of these radical expressions areunits(a unit is an algebraic integer α such that αβ=1 where β is another algebraic integer).For me the only way to establish these identities is to raise each side of the equation to an appropriate power using brute force algebra and then check the equality. However for higher powers (for example equation (2) above) this seems very difficult.

Is there any underlying structure in these powers of units which gives rise to such identities or these are mere strange cases which were noticed by Ramanujan who used to play with all sorts of numbers as a sort of hobby? I believe (though not certain) that perhaps Ramanujan did have some idea of such structure which leads to some really nice relationships between units and their powers. I wonder if there is any sound theory of such relationships which can be exploited to give many such identities between nested and denested radicals.

**Answer**

We also have the following identity,

3√m3−n3+6m2n+3mn2−3(m2+mn+n2)3√mn(m+n)=3√m2(m+n)−3√mn2−3√(m+n)2n

For m=n=1 we get (1).

For m=4 and n=1 we get (5).

**Attribution***Source : Link , Question Author : Paramanand Singh , Answer Author : Tito Piezas III*