# 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

• What is unsatisfactory about writing it as $\sqrt{\sqrt{4}+\sqrt{16}+1}$? – JMoravitz Sep 19 '17 at 16:08
• @MrYouMath thanks – Vindhyachal Singh Sep 19 '17 at 16:13
• I have mentioned the answer .i need a process to arrive there – Vindhyachal Singh Sep 19 '17 at 16:24

## 4 Answers

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

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

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 $\sqrt{2}$ 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 $1+\sqrt{2}$

• thank you very much. ...I am interested in answer 😂 – Vindhyachal Singh Sep 19 '17 at 16:30

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

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

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