Square root of radicals Find the square root of $4^{1/3}+16^{1/3}+1$.
I tried to solve by supposing the square root to be $x$ and then cubing both sides but it didn't work. 
I do not need exact value. 
By hit and trial
I have seen that answer should involve $2^{1/3}+1....$ 
Thanks
I will assume you are interested only in real numbers and not complex numbers.  If you wish to simplify so that there are no nested radicals:

First express inner radicals as exponentials terms of the same base

$\sqrt{1+\sqrt[3]{4}+\sqrt[3]{16}}=\sqrt{1+2^{2/3}+2^{4/3}}=\sqrt{1+2\cdot 2^{1/3}+2^{2/3}}$

Above, we used that $2^{4/3}=2^{3/3}\cdot 2^{1/3}$.

Now, let $x=2^{1/3}$.  We recognize that the above can be written as

$\sqrt{1+2x+x^2}=\sqrt{(1+x)^2}=|1+x|$.  Remembering that $\sqrt[3]{2}$ is a positive real, we may replace this back in for $x$ and remove the absolute value sign to get that the original expression is equal to $1+\sqrt[3]{2}$
$1^{2/3},2^{2/3},4^{2/3}$ is GP with $a=1^{2/3}, r=2^{2/3}, n=3$

$\therefore S=1^{2/3}\cdot \frac{(2^{2/3})^3-1}{2^{2/3}-1}=\frac{3}{2^{2/3}-1}$

$\therefore$ the required answer$=\sqrt{\frac{3}{2^{2/3}-1}}$