Square root of radicals

Question

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

Answer 1

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 $1+\sqrt[3]{2}$

Answer 2

Notice that $4^{1/3}=2^{2/3}=\left(2^{1/3}\right)^2$ and $16^{1/3}=(8\cdot2)^{1/3}=2\cdot2^{1/3}$. Therefore: $$4^{1/3}+16^{1/3}+1=\left(2^{1/3}\right)^2+2\cdot2^{1/3}\cdot1+1^2=\left(2^{1/3}+1\right)^2.$$

Answer 3

$1^{2/3},2^{2/3},4^{2/3}$ 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$=\sqrt{\frac{3}{2^{2/3}-1}}$

Answer 4

$$16^{1/3}=2^{1/3} . 8^{1/3}=2 . 2^{1/3}$$ So we have $$2^{{1/3}^2} +2 . 2^{1/3}+1 = (2^{1/3}+1)^2$$ So the root is $$2^{1/3} +1$$