Denesting of Ramanujan's radicals

Note that

$\begin{aligned} (\sqrt 3 + \sqrt 2)^4 & = (\sqrt 3)^4 + 4 (\sqrt 3)^3\sqrt 2 + 6 (\sqrt 3)^2(\sqrt 2)^2 + 4 \sqrt 3(\sqrt 2)^3 + (\sqrt 2)^4 \\ & = 9 + 12 \sqrt 6 + 36 + 8 \sqrt 6 + 4 \\ & = 49 + 20 \sqrt 6 \end{aligned}$

Similarly, $\begin{aligned} (\sqrt 3 - \sqrt 2)^4 & = 49 - 20 \sqrt 6 \end{aligned}$ . And

$\begin{aligned} \sqrt [4]{49 + 20 \sqrt 6} +\sqrt [4]{49 - 20 \sqrt 6} & = \sqrt [4]{(\sqrt 3 + \sqrt 2)^4} +\sqrt [4]{(\sqrt 3 - \sqrt 2)^4} \\ & = \sqrt 3 + \sqrt 2 + \sqrt 3 - \sqrt 2 \\ & = 2\sqrt 3 \end{aligned}$

$\implies a+b = 2 + 3 = \boxed{5}$

$(x + y)^{4} = x^{4} + 4x^{3}y + 6x^{2}y^{2} + 4xy^{3} + y^{4} = x^{4} + y^{4} + 4xy(x^{2} + y^{2}) + 6(xy)^{2}$

$\Longrightarrow (x + y)^{4} = x^{4} + y^{4} + 4xy((x + y)^{2} - 2xy) + 6(xy)^{2} = x^{4} + y^{4} + 4xy(x + y)^{2} - 2(xy)^{2}.$

Now if $\large x = \sqrt[4]{49 + 20\sqrt{6}}$ and $\large y = \sqrt[4]{49 - 20\sqrt{6}}$ we have that

$x^{4} + y^{4} = (49 + 20\sqrt{6}) + (49 - 20\sqrt{6}) = 98$ and $xy = \sqrt[4]{2401 - 400*6} = 1.$

Thus $(x + y)^{4} = 98 + 4(x + y)^{2} - 2$

$\Longrightarrow (x + y)^{4} - 4(x + y)^{2} - 96 = 0$

$\Longrightarrow ((x + y)^{2} - 12)((x + y)^{2} + 8) = 0,$

so either $(x + y)^{2} = 12$ or $(x + y)^{2} = -8.$

assuming $x$ and $y$ to be real numbers, we can say that $(x + y)^{2} = 12,$ and so $x + y = \sqrt{12} = 3\sqrt{2}$

Hence $a + b = \boxed{5}.$

√(√(49±20√6)) = √(5 ±2√6), and √(5 ±2√6) = √3 ± √2,

So   ∜(49+20√6)  + ∜(49-20√6)  = 2√3, and  a+b = 5