How to simplify the cubic radical $\sqrt[3]{a\sqrt{b}-c}$?

Question

How to simplify the cubic radical $\sqrt[3]{a\sqrt{b}-c}$ ?

So I encountered a particular problem in chapter of surds and radicals to find the cube root of $38\sqrt{14}-100\sqrt{2}$ .

So I took out 2√2 common leaving to find out the cube root of $\sqrt[3]{19\sqrt{7}-50}$

From I'm having no good idea to continue further ..

Answer 1

If there is a nice solution, which I will here take to mean integral, then it must be that $-50+19\sqrt 7=(a+b\sqrt7)^3$ with nice $a,b$. There are several things we can tell from this, most immediately that

• $a^3+21ab^2=-50$
• $3a^2b+7b^3=19$

If $a$ and $b$ really are nice, then the first line tells us that $a$ is negative and a divisor of $50$ (since $a^2+21b^2$ is necessarily positive). Similarly we find that $b$ is positive and a divisor of $19$.

That makes it a total of twelve possibilities to check. Considering that $3a^2+7b^2$ is also a divisor of $19$, that narrows it down a lot further.

Answer 2

Remember that we have: $$ (a-b) ^3 = a^3 -3a^2 b + 3 a b^2 - b^3.$$

You're already very close to the answer. First you should notice that your formula is already of the form of $(x-y)$, and so the first thing you should think of is how can we rewrite our result into equation 1.

We're trying to find $a$ and $b$ such that $(a-b) ^3 = 19\sqrt{7} - 50.$

Another hint is that $ \sqrt{7}^3 = 7\sqrt{7} $ and that $2^3 = 8.$

Finally you'll obtain : $(\sqrt{7}-2)^3 = 19\sqrt{7} - 50.$

Answer 3

So , the idea is $\sqrt[3]{19\sqrt{7}-50}$ will be of the form $a\sqrt{7}-b$ . Therefore,

Let $\sqrt[3]{19\sqrt{7}-50} = a\sqrt{7}-b$ .

Then, $\sqrt[3]{19\sqrt{7}+50} = a\sqrt{7}+b$

By multiplying , $3 = 7a^2-b^2$ ...(1)

And $19\sqrt{7}-50=7\sqrt{7}a^3-21a^2b+3\sqrt{7}ab^2-b^3$

whence, $50 = 21a^2b+b^3$ ...(2)

From (1) and (2), $b = 2$

Hence, $a=1$, $b=2$ ($a$ should be obviously positive in order to make the cube root positive)
$$\therefore \sqrt[3]{19\sqrt{7}-50} = \sqrt{7}-2$$

Answer 4

OP's $\;\sqrt[3]{19\sqrt{7}-50}\;$ is one of the cases covered by my answer here.

Therefore a sufficient condition for $\,a,b = \sqrt[3]{m \sqrt{p} \pm n\sqrt{q}}\,$ to denest is for $\,m^2 \cdot p - n^2 \cdot q\,$ to be the cube of a rational $\,r\,$, and for the cubic $\,p\, t'^{\,3} - 3r\, t' - 2m\,$ to have an appropriate rational root, and in that case $\,a,b = \frac{1}{2}\left(t'\,\sqrt{p} \pm \sqrt{t'^{\,2} p-4r}\right)\,$.

For $\,a,b = \sqrt[3]{19\sqrt{7}\pm 50}\,$:

• $m=19\,$, $\,p=7\,$, $\,n=\pm 50\,$, $\,q=1\,$;

• $m^2 \cdot p - n^2 \cdot q$ $= 19^2 \cdot 7 - (\pm 50)^2 \cdot 1$ $= 27$ $=3^3$ $\implies r=3\,$;

• $0 = p\, t'^{\,3} - 3r\, t' - 2m = 7 t'^{\,3} - 9 t' - 38\,$ with the only real root $\,t' = 2\,$.

Then:

$$ a,b = \frac{1}{2}\left(t'\,\sqrt{p} \pm \sqrt{t'^{\,2} p - 4r}\right) = \frac{1}{2}\left(2 \cdot \sqrt{7} \pm \sqrt{2^2 \cdot 7 - 4 \cdot 3}\right) = \frac{1}{2}\left(2\sqrt{7} \pm \sqrt{16}\right) = \sqrt{7} \pm 2$$