I have recently came across a problem involving cube roots of numbers added to square roots, in the form of $\sqrt[3]{X+Y\sqrt{Z}}$, this is called a “nested radical.” I already figured out a nested square root, I decided to give this a go.

I can take a number like $(1+\sqrt{5}{)}^2$ and work out it is equal to $16+8\sqrt{5}$. Unfortunately, it is very difficult to recognize that $1+\sqrt{5}$ is the cube root (or even a factor) of that. Even harder, trying to find an equation such as $\sqrt[3]{1080-624\sqrt{3}}$.

Recently, I developed a method of breaking down the equation into its cube root equation.

With an equation in the form of $\sqrt[3]{X+Y\sqrt{Z}}$, where Z contains no perfect square factors.

Let’s start with
$$\begin{array}{rl}\sqrt[3]{X+Y\sqrt{Z}}& =A+B\sqrt{Z}\\ X+Y\sqrt{Z}& =(A+B\sqrt{Z}{)}^3\\ & =A^3+3A^{2}B\sqrt{Z}+3A{B}^{2}Z+B^{3}Z\sqrt{Z}\\ & =A^3+3A{B}^{2}Z+3A^{2}B\sqrt{Z}+B^{3}Z\sqrt{Z}\\ & =A^3+3A{B}^{2}Z+(3A^{2}B+B^{3}Z)\sqrt{Z}.\end{array}$$

We need to find
$$\begin{array}{}\text{(1)}& X& =A^3+3A{B}^{2}Z\text{(2)}& Y& =3A^{2}B+B^{3}Z.\end{array}$$

If You look at X, You see it is the sum of a cube and a multiple of 3Z.

Let’s look at an example, $\sqrt[3]{1080-624\sqrt{3}}$. If we subtract cubes from the initial number(1080) we get:

$$\begin{array}{rlr}1080-1^3& =1080-1=1079& No\\ 1080-2^3& =1080-8=1072& No\\ 1080-3^3& =1080-27=1053=3Z\cdot 117& Yes\\ 1080-4^3& =1080-64=1016& No\\ 1080-5^3& =1080-125=955& No\\ 1080-6^3& =1080-216=864=3Z\cdot 96& Yes\\ 1080-7^3& =1080-343=737& No\\ 1080-8^3& =1080-512=568& No\\ 1080-9^3& =1080-729=351=3Z\cdot 39& Yes\\ 1080-{10}^3& =1080-1000=80& No\\ 1080-{11}^3& =1080-1331=-251& No\end{array}$$

We need a result that is evenly divisible by 3Z, or 9 in this case. ${11}^3$ is greater than 1080, so we stop. (Even if all answers are complex, B² will still be Real.)

Our results show 3, 6, and 9 are viable solutions for a, we must now check each against X to get B.

if(A=3)	$1053=9\cdot 3\cdot B^2$	B²=39	$B=\sqrt{39}$
if(A=6)	$864=9\cdot 6\cdot B^2$	B²=16	$B=4$
if(A=9)	$351=9\cdot 9\cdot B^2$	$B^2=4\frac{1}{3}$	$B=\sqrt{\frac{13}{3}}$

The only even square root is for A=6, B=4. $X=6^3+3\cdot 6\cdot 4^2\cdot 3=1080$, so this is a possible solution pair. First we must now check against Y.

$$\begin{array}{rl}Y& =3A^{2}B+B^{3}Z\\ & =3\cdot 6^2\cdot 4+4^3\cdot 3\\ & =3\cdot 36\cdot 4+64\cdot 3\\ & =432+192\\ & =624\end{array}$$

It matches, so we have a correct solution for $\sqrt[3]{1080-624\sqrt{3}}=6-4\sqrt{3}$. Remember, as the original had a negative Y, the solution must negate B.

If our example $\sqrt[3]{1080-624\sqrt{3}}$ was actually $\sqrt[3]{1080-623\sqrt{3}}$, then the formulas above would not work (there would not be any solution) and would not be able to be simplified any further.

 

Let’s try $\sqrt[3]{\frac{371}{125}+\frac{319}{100}\sqrt{2}}$, which can also be written $\sqrt[3]{2.968+3.19\sqrt{2}}$. This is a bit harder as we are not dealing with integers. It may be easier to try smaller cubes.

$$\begin{array}{rl}2.968-{.1}^3& =2.968-.001=2.967=3Z\cdot 0.4945\\ 2.968-{.2}^3& =2.968-.008=2.960\\ 2.968-{.3}^3& =2.968-.027=2.941\\ 2.968-{.4}^3& =2.968-.064=2.904=3Z\cdot 0.4840\\ 2.968-{.5}^3& =2.968-.125=2.843\\ 2.968-{.6}^3& =2.968-.216=2.752\\ 2.968-{.7}^3& =2.968-.343=2.625=3Z\cdot 0.4375\\ 2.968-{.8}^3& =2.968-.512=2.456\\ 2.968-{.9}^3& =2.968-.729=2.239\\ 2.968-{1.0}^3& =2.968-1.000=1.968=3Z\cdot 0.328\\ 2.968-{1.1}^3& =2.968-1.331=1.637\\ 2.968-{1.2}^3& =2.968-1.728=1.240\\ 2.968-{1.3}^3& =2.968-2.197=0.771=3Z\cdot 0.1285\\ 2.968-{1.4}^3& =2.968-2.744=0.224\\ 2.968-{1.5}^3& =2.968-3.375=-0.407\end{array}$$

Of these, 0.1, 0.4, 0.7, 1, and 1.3 are finitely divisible by 6 (3Z). If we didn’t find any, we would have to try smaller numbers.

if (A=0.1)	$2.967=6\cdot 0.1\cdot B^2$	$B^2=\frac{4945}{1000}$	B is irrational
if (A=0.4)	$2.904=6\cdot 0.4\cdot B^2$	$B^2=\frac{121}{100}$	$B=\frac{11}{10}$
if (A=0.7)	$2.625=6\cdot 0.7\cdot B^2$	$B^2=\frac{5}{8}$	$B=\sqrt{\frac{5}{8}}$
if (A=1)	$1.968=6\cdot 1\cdot B^2$	$B^2=\frac{41}{125}$	B is irrational
if (A=1.3)	$0.771=6\cdot 1.3\cdot B^2$	$B^2=\frac{257}{2600}$	B is irrational

Three of these are obviously irrational, so cannot be part of our solution, so are not calculated.

For A=0.7, B is also irrational, so we will first test A=0.4:

$$\begin{array}{rl}2.968& =A^3+3A{B}^{2}Z\\ & ={0.4}^3+3\cdot 0.4\cdot B^2\cdot 2\\ & =0.064+2.4\cdot B^2\\ 2.904& =2.4B^2\\ 1.21& =B^2\\ B& =\frac{11}{10}=1.1\end{array}$$

And for Y:

$$\begin{array}{rl}Y& =3A^{2}B+B^{3}Z\\ & =3\cdot {0.4}^2\cdot 1.1+{1.1}^3\cdot 2\\ & =3\cdot 0.16\cdot 1.1+1.331\cdot 2\\ & =0.528+2.662\\ & =3.19\end{array}$$

It matches, so we have a correct solution for $\sqrt[3]{2.968+3.19\sqrt{2}}=0.4+1.1\sqrt{2}$.

Any cube root in this form must also have $X^2+Y^2\cdot Z$ equal a perfect cube.

$\sqrt[3]{{1080}^2-{624}^2\cdot 3}=-12$ and $\sqrt[3]{{\frac{371}{125}}^2-{\frac{319}{100}}^2\cdot 2}=\frac{-113}{50}$, so these are solvable.

Also, remember that these are cube roots, so there are two complex roots. Just rotate the answer found by 120° and 240°, for the remaining answers. If this technique doesn’t find a simple answer, then all 3 may be complex.

---

Another way to solve, if Z=5 and the number under the radical is in the form of $2K\pm K\sqrt{5}$, aka $K\cdot (2\pm \sqrt{5})$, the answer will be in the form of $A\pm +A\sqrt{5}$. In this case, $A=\sqrt[3]{\frac{K}{8}}$.

Alternate values for Z:

Z	Form	A=
2	$K(7\pm 5\sqrt{2})$	$\sqrt[3]{K}$
3	$K(5\pm 3\sqrt{3})$	$\sqrt[3]{\frac{K}{2}}$
5	$K(2\pm \sqrt{5})$	$\sqrt[3]{\frac{K}{8}}$
6	$K(19\pm 9\sqrt{6})$	$\sqrt[3]{K}$
7	$K(11\pm 5\sqrt{7})$	$\sqrt[3]{\frac{K}{2}}$
8	$K(25\pm 11\sqrt{8})$ or $K(25\pm 22\sqrt{2})$	$\sqrt[3]{K}$
10	$K(31\pm 13\sqrt{10})$	$\sqrt[3]{K}$

This is likely not the best method for finding the cube roots, but it does work.