Nested roots, part 2

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 $$(1+\sqrt5)^2$$ and work out it is equal to $$16 + 8\sqrt5$$. Unfortunately, it is very difficult to recognize that $$1+\sqrt5$$ is the cube root (or even a factor) of that. Even harder, trying to find an equation such as $$\sqrt[3]{1080-624\sqrt3}$$.

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.

\begin{align} \sqrt[3]{X+Y\sqrt{Z}} &= A+B\sqrt{Z} \\ X+Y\sqrt{Z} &= (A+B\sqrt{Z})^3 \\ &=A^3 + 3A^2B\sqrt{Z} + 3AB^2Z + B^3Z\sqrt{Z} \\ &=A^3 + 3AB^2Z + 3A^2B\sqrt{Z} + B^3Z\sqrt{Z} \\ &=\color{Yellow}{A^3 + 3AB^2Z} +\color{cyan}{( 3A^2B + B^3Z})\sqrt{Z} .\\ \end{align}

We need to find
\begin{align} X &= \color{Yellow}{A^3 + 3AB^2Z} \tag{1}\label{1} \\ Y &=\color{cyan}{3A^2B + B^3Z}. \tag{2}\label{2}\\ \end{align}

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

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, 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) $$351 =9\cdot 9\cdot B^2$$ $$B^2=4\frac13$$ $$B=\sqrt{\frac{13}{3}}$$

The only even square root is for A=6, B=4. $$X= 6^3 + 3\cdot6 \cdot 4^2 \cdot 3 =1080$$, so this is a possible solution pair. First we must now check against Y.

\begin{align} Y &=3A^2B + B^3Z\\ &=3\cdot 6^2\cdot4 + 4^3\cdot3\\ &=3\cdot 36\cdot4 + 64\cdot3\\ &=432 + 192\\ &=624\\ \end{align}

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 $$\sqrt[3]{\frac{371}{125}+\frac{319}{100}\sqrt2 }$$, which can also be written $$\sqrt[3]{2.968+3.19\sqrt2 }$$. This is a bit harder as we are not dealing with integers. It may be easier to try smaller cubes.

\begin{align} 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{align}

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) $$2.625=6\cdot 0.7\cdot B^2$$ $$B^2=\frac58$$ $$B=\sqrt{\frac58}$$ 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{align} 2.968 &= A^3 + 3AB^2Z\\ &=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{align}

And for Y:

\begin{align} Y &=3A^2B + B^3Z\\ &=3\cdot 0.4^2\cdot1.1 + 1.1^3\cdot2\\ &=3\cdot 0.16\cdot1.1 + 1.331\cdot2\\ &=0.528 + 2.662\\ &=3.19\\ \end{align}

It matches, so we have a correct solution for $$\sqrt[3]{2.968+3.19\sqrt2 } =0.4+1.1\sqrt{2}$$.

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 $$\sqrt[3]{\frac{371}{125}^2-\frac{319}{100}^2\cdot2}=\frac{-113}{50}$$, 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 $$2K\pm K\sqrt5$$, aka $$K\cdot(2\pm \sqrt5)$$, the answer will be in the form of $$A\pm +A\sqrt5$$. In this case, $$A=\sqrt[3]{\frac{K}{8}}$$.

Alternate values for Z:

 Z Form A= 2 $$K(7\pm 5\sqrt2)$$ $$\sqrt[3]{K}$$ 3 $$K(5\pm 3\sqrt3)$$ $$\sqrt[3]{\frac{K}{2}}$$ 5 $$K(2\pm \sqrt5)$$ $$\sqrt[3]{\frac{K}{8}}$$ 6 $$K(19\pm 9\sqrt6)$$ $$\sqrt[3]{K}$$ 7 $$K(11\pm 5\sqrt7)$$ $$\sqrt[3]{\frac{K}{2}}$$ 8 $$K(25\pm 11\sqrt8)$$ or $$K(25\pm 22\sqrt2)$$ $$\sqrt[3]{K}$$ 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.

Factoring numbers with square root terms

Quite often, when doing calculations on polyhedra, you will find yourself with complex equations like $$\frac{20+7\sqrt3}{1+2\sqrt3}$$.

Is the top evenly divisible by the bottom? I certainly can’t tell just by looking at them.

There must be a method to factor the numerator.

Quite often, when doing calculations on polyhedra, you will find yourself with complex equations like $$$$\tag{eq1}\label{eq1}\frac{20+7\sqrt3}{1+2\sqrt3}$$$$

Is the top evenly divisible by the bottom? I certainly can’t tell just by looking at them.

There must be a method to factor the numerator.

Let’s start with $$\color{blue}{(A+B\sqrt{R})}\color{lime}{(C+D\sqrt{R})}$$ multiplying the terms together gives $$\color{salmon}{AC+BRD+(BC+AD)\sqrt{R}}$$

So, $$\frac{\color{salmon}{AC+BRD+(BC+AD)\sqrt{R}}}{\color{blue}{A+B\sqrt{R}}} = \color{lime}{C+D\sqrt{R}}$$

Then we can substitute known values (A=1, B=2, R=3) into the formula. $$\frac{1C+2(3)D+(2C+1D)\sqrt{3}}{1+2\sqrt{3}} = C+D\sqrt{3}$$

Finding the answer is just a matter of finding C and D.

Consider that
$$\frac{\color{yellow}{C+6D} + \color{cyan}{(2C+D)}\sqrt3}{1+2\sqrt3} = \frac{\color{yellow}{20}+\color{cyan}{7}\sqrt3}{1+2\sqrt3}$$

We can break the equation into the colored sections to find \begin{align}\color{yellow}{C+6D} & \color{yellow}{=20} \\ C&=20-6D\tag{eq2}\label{eq2} \end{align} and $$\color{cyan}{2C+D = 7}\tag{eq3}\label{eq3}$$

We now know C \eqref{eq2}, at least in terms of D, then substitute this into \eqref{eq3}.
\begin{align}2(20-6D)+D &= 7 \\ 40-12D+D &= 7 \\ -11D &= 7-40 = -33 \\D &= \frac{-33}{-11} \\ D&=3\end{align}

Returning to \eqref{eq2} and substituting D=3, we get $$C = 20-6\cdot 3 = 20-18 = 2$$.

Then our answer should be $$2+3\sqrt{3}$$, and it is, because $$(1+2\sqrt3)(2+3\sqrt3) = 20+7\sqrt3$$.

This method make it simple to divide out these complicated numbers, although, you might not get a simpler answer. For example, if we change \eqref{eq1} to $$\Large\frac{19+7\sqrt3}{1+2\sqrt3}$$, we end up with answers C=23/11 and D=31/11, but it still works as $$(1+2\sqrt3)(\frac{23}{11}+\frac{31\sqrt3}{11}) = 19+7\sqrt3$$. Although, we could write our answer as $$\frac{1}{11}(23+31\sqrt3)$$.

Another example:$$\frac{-3}{3+2\sqrt3}$$
We have values (A=3, B=2, R=3), giving $$3C + 2\cdot3D + (2C+3D)\sqrt3 = (3+2\sqrt3)(C+D\sqrt3)$$

Remember that $$-3$$ is the same as $$-3+0\sqrt3$$. So
\begin{align}3C+6D&=-3\label{e2}\tag{eq4}\\ 2C+3D&=0\label{e3}\tag{eq5} \end{align}

Working out \eqref{e2} gives
\begin{align}3C+6D&=-3 \\ 6D&= -3-3C \\D&=\frac{3(-1-C)}{6} \\D&= \frac{-1-C}{2}\end{align}

Plugging into \eqref{e3}, gives
\begin{align}2C+3D&=0 \\ 2C+3\cdot(\frac{-1-C}{2})&=0 \\ \frac{4C}{2}+\frac{-3C-3}{2}&=0 \\ \frac{4C-3C-3}{2}&=0 \\ \frac{C-3}{2}&=0 \\ C-3&=0 \\C&=3\end{align}

With C=3, we can find D in \eqref{e2} \begin{align}3(3)+6D&=-3 \\ 9+6D&=-3 \\ 6D&=-3-9 \\ 6D&=-12 \\ D&=\frac{-12}{6} \\ D&=-2\end{align}

So the answer is $$3-2\sqrt3$$, as $$(3+2\sqrt3)(3-2\sqrt3)=-3$$.

You might notice this is the same as the “Difference of 2 squares” (x-y)(x+y) = x²-xy+xy-y² = x²-y². The “xy” terms cancel each other out, leaving 0. (x-y) and (x+y) are conjugates of one another, so when you have a numerator with a zero times the square root (or in other words, the numerator is a whole number), you may wish to try the denominator’s conjugate first. It may not be the answer you seek, but it can help you sometimes.

If the denominator is a whole number (or rational), $$\frac{6+2\sqrt3}{-2}$$

then divide both parts of the numerator by the denominator,

$$\frac{6}{-2}+\frac{2\sqrt3}{-2}=-3-\sqrt3$$

So to find an answer to $$\notag\frac{X+Y\sqrt{R}}{A+B\sqrt{R}} = C+D\sqrt{R}$$

You will need to find the solutions to $$AC+BRD=X$$ and $$BC+AD=Y$$

UPDATE: I have also realized that you can just multiply the top and bottom of the fraction by the denominator’s conjugate.

In the first equation:

\begin{align}\frac{20+7\sqrt3}{1+2\sqrt3}&=\frac{(20+7\sqrt3)(1-2\sqrt3)}{(1+2\sqrt3)(1-2\sqrt3)}\\&=\frac{20-42+7\sqrt3-40\sqrt3}{1-12}\\&=\frac{-22-33\sqrt3}{-11}\\&=2+3\sqrt3\end{align}

In this method, the denominator is always a whole number (if both A&B are whole numbers).

sqr(A+B*sqr(C))

Lately, I have been doing a lot of math involving square roots of numbers added to square roots, in the form of $$\sqrt{A+B\sqrt{C}}$$, this is called a “nested radical.” Normally, you would not be able to simplify any further, unless there was a common factor of both that could be removed, or if both items had the same number under the radical (eg. $$\sqrt{4\sqrt7+12\sqrt7}=\sqrt{16\sqrt7}=4\sqrt[4]{7}$$ ).

I can take a number like $$(1+\sqrt5)^2$$ and work out it is equal to $$6 + 2\sqrt5$$. Unfortunately, it is very difficult to recognize that $$1+\sqrt5$$ is the square root (or even a factor) of that. Even harder, trying to find an equation such as $$\sqrt{469+144\sqrt5}$$.

Recently, I found a method of breaking down the equation into its square root equation.

With an equation in the form of $$\sqrt{A+B\sqrt{C}}$$, where C contains no perfect square factors, taking the formula $$D^2 = A^2 – (B^2)*C$$, if D is a rational number, you will be able to factor the equation, otherwise it is simplified as much as is possible.

Then you will be able to get the formula: $$\sqrt{A+B\sqrt{C}} = \pm \left(\sqrt{\frac{A-D}{2}} + sign(A*B)*\sqrt{\frac{A+D}{2}}\right)$$, where sign(A*B) is the positive or negative value of A*B, it will always give you 0, +1, or -1.

So to find $$\sqrt{469+144\sqrt5}$$, we find $$469^2 – 144^2 * 5 = 116281$$. Since 116281 = 341*341, we can plug D=341 into the formula to get:

\begin{align} \sqrt{\frac{469-341}{2}} + 1*\sqrt{\frac{469+341}{2}}&=\sqrt{\frac{128}{2}} + \sqrt{\frac{810}{2}}\\ &=\sqrt{64} + \sqrt{405} \\ &= 8 + \sqrt{81*5} \\ &= 8+9\sqrt{5}\end{align}

So, $$\sqrt{469+144\sqrt5} = (8+9\sqrt{5}) \text{ and } (-8-9\sqrt{5})$$.

To find $$\sqrt{53-12\sqrt{11}}$$, we find $$53^2 – 12^2 * 11 = 1225$$. Since 1225 = 35*35, we can plug D=35 into the formula to get:

\begin{align*} \sqrt{\frac{53-35}{2}} + -1*\sqrt{\frac{53+35}{2}} &=\sqrt{\frac{18}{2}} – \sqrt{\frac{88}{2}}\\ &=\sqrt{9} – \sqrt{44} \\ &= 3 – \sqrt{4*11} \\ &= 3-2\sqrt{11} \end{align*}

So, $$\sqrt{53-12\sqrt11} = (3-2\sqrt{11}) \text{ and } (-3+2\sqrt{11})$$.

If A and B have the same sign, the answer will have both terms the same sign ({+,+} and {-,-} are equally valid). If A and B have different signs, the terms in the answer will have different signs ({+,-} and {-,+}). Often you will need to determine the complete answer that equates to positive, after all, it is hard to have a real object with a negative length.

An interesting correlation: $$(X-Y)^2 = (Y-X)^2$$, but remember $$\sqrt{X-Y} \not= \sqrt{Y-X}$$. Double check your signs by re-squaring your answer.

Note: This still works if D is not an integer, but is rational:

\begin{align*} \sqrt{8.41+2.64\sqrt5} &=\sqrt{\frac{8.41-5.99}{2}} + \sqrt{\frac{8.41+5.99}{2}} &D=\sqrt{8.41^2 + 2.64^2\cdot5}=5.99\\ &=\sqrt{\frac{2.42}{2}} + \sqrt{\frac{14.4}{2}}\\ &=\sqrt{\frac{121}{100}} + \sqrt{\frac{720}{100}}\\ &=\frac{11}{10} + \frac{12\sqrt5}{10}\\ &=1.1 + 1.2\sqrt5\\ \end{align*}

Which is correct. $$\left[1.1+1.2\sqrt5\right]^2=8.41+2.64\sqrt5$$