An analogue to Ramanujan's identity $\left\{3\left(\left(a^{3}+b^{3}\right)^{1/3}-a\right)\left(\left(a^{3}+b^{3}\right)^{1/3}-b\right)\right\}^{1/3}$

Question

In this paper discussing Ramanujan's submitted questions to the Mathematical journal, Question 785 states:

Show that; $$\small{\left\{3\left(\left(a^{3}+b^{3}\right)^{1/3}-a\right)\left(\left(a^{3}+b^{3}\right)^{1/3}-b\right)\right\}^{1/3}=\left(a+b\right)^{2/3}-\left(a^{2}+b^{2}-ab\right)^{1/3}\tag{1}}$$ This is anagolous to, $$\small{\left\{2\left(\left(a^{2}+b^{2}\right)^{1/2}-a\right)\left(\left(a^{2}+b^{2}\right)^{1/2}-b\right)\right\}^{1/2}=\left(a+b\right)-\left(a^{2}+b^{2}\right)^{1/2}\tag{2}}$$

My enquiry is whether an analogue of Ramanujan's analogue is possible but with the forth-power? That is a closed form (aka a simplified form) for:

$$\left\{4\left(\left(a^{4}+b^{4}\right)^{1/4}-a\right)\left(\left(a^{4}+b^{4}\right)^{1/4}-b\right)\right\}^{1/4}=f\left(a,b\right)\tag{3}$$

Is this feasible via elementary factorizing?

Answer 1

Let $a^2+b^2=2kab$.

Thus, we need to prove that $$3\left(\sqrt[3]{(a+b)^2(a^2-ab+b^2)^2}-\sqrt[3]{(a+b)^4(a^2-ab+b^2)}+ab\right)=$$ $$=\left(\sqrt[3]{a^2+b^2+2ab}-\sqrt[3]{a^2+b^2-ab}\right)^3$$ or $$3\left(\sqrt[3]{(2k+2)(2k-1)^2}-\sqrt[3]{(2k+2)^2(2k-1)}+1\right)=$$ $$=\left(\sqrt[3]{2k+2}-\sqrt[3]{2k-1}\right)^3,$$ which is true because $$2k+2-(2k-1)=3.$$ We can get a solution of the second problem by squaring of the both sides.

About your third problem.

I think it's impossible to find here something nice as Ramanujan's identities.

Answer 2

There is an identity involving fourth powers shown below:

$2[(a^4+b^4)-(a^2+b^2)(a^4+b^4)^{1/2}+(ab)^2]=[(a^2+b^2)-(a^4+b^4)^{1/2}]^{2}$

And also for eight powers:

$2[(a^8+b^8)-(a^4+b^4)(a^8+b^8)^{1/2}+(ab)^4]=[(a^4+b^4)-(a^8+b^8)^{1/2}]^{2}$