Getting rid of cube roots in the form of (a+b)+(a-b)

Question

So, I've come across a question that asked "Simplify the sum $\sqrt[3]{18+5\sqrt{13}}+\sqrt[3]{18-5\sqrt{13}}$". As I've had little to no experience with these kinds of questions, I would really appreciate it if someone answers this question and provides a way to remove double cube roots like this. I've tried $\sqrt[3]{18+5\sqrt{13}}+\sqrt[3]{18-5\sqrt{13}}=x$ and cubing both sides but that just made it even messier. I thought about multiplying both sides by $\sqrt[3]{(18+5√13)}^2$ but abandoned the attempt after realising I gave x a complicated coefficient.

Answer 1

Since $(a+b)^3=a^3+b^3 + 3ab(a+b)$

$(\sqrt[3]{18+5\sqrt{13}}+\sqrt[3]{18-5\sqrt{13}})^3=18+5\sqrt{13} + 18-5\sqrt{13} +3\times \sqrt[3]{18+5\sqrt{13}}\times \sqrt[3]{18-5\sqrt{13}}(\sqrt[3]{18+5\sqrt{13}}+\sqrt[3]{18-5\sqrt{13}}) $

$(\sqrt[3]{18+5\sqrt{13}}+\sqrt[3]{18-5\sqrt{13}})^3=18+5\sqrt{13} + 18-5\sqrt{13} +3\times \sqrt[3]{(18+5\sqrt{13})\times (18-5\sqrt{13})}(\sqrt[3]{18+5\sqrt{13}}+\sqrt[3]{18-5\sqrt{13}}) $

$(\sqrt[3]{18+5\sqrt{13}}+\sqrt[3]{18-5\sqrt{13}})^3=18+5\sqrt{13} + 18-5\sqrt{13} +3\times \sqrt[3]{-1}(\sqrt[3]{18+5\sqrt{13}}+\sqrt[3]{18-5\sqrt{13}}) $

$(\sqrt[3]{18+5\sqrt{13}}+\sqrt[3]{18-5\sqrt{13}})^3=36 +3\times (-1)\times (\sqrt[3]{18+5\sqrt{13}}+\sqrt[3]{18-5\sqrt{13}}) $

Let $x=\sqrt[3]{18+5\sqrt{13}}+\sqrt[3]{18-5\sqrt{13}}$

$x^3+3x-36=(x-3)(x^2 +3x+12)=0$

$x=3$ is the only real solution

Answer 2

Analytically compute $\displaystyle ~\sqrt[3]{18 + 5\sqrt{13}} + \sqrt[3]{18 - 5\sqrt{13}}.$

$\underline{\text{Preliminary Considerations}}$

$$\left[ ~r + s\sqrt{13} ~\right]^3 = \left[r^3 + (39 rs^2) \right] + \sqrt{13}\left[ ~3r^2s + 13s^3 ~\right]. \tag1 $$

$$\left[ ~r - s\sqrt{13} ~\right]^3 = \left[r^3 + (39 rs^2) \right] - \sqrt{13}\left[ ~3r^2s + 13s^3 ~\right]. \tag2 $$

So, you are looking for real numbers $~r,s,~$ such that :

• $~\displaystyle \left[ ~r + s\sqrt{13} ~\right]^3 = 18 + 5\sqrt{13}.$
• $~\displaystyle \left[ ~r - s\sqrt{13} ~\right]^3 = 18 - 5\sqrt{13}.$

I know of two generic approaches:

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$\underline{\text{Method 1}}$

Using a calculator, you have that

$$\sqrt[3]{18 + 5\sqrt{13}} + \sqrt[3]{18 - 5\sqrt{13}} = 3.$$

This implies that $~r = \dfrac{3}{2}.~$

Then, the next step is to use the first term on the RHS of (1) above to solve for $~s~$ where

$$\frac{27}{8} + \left[ ~\frac{39 \times 3}{2} s^2 ~\right] = 18 \implies $$

$$\frac{117}{8} = \frac{117}{2} s^2 \implies s = \pm \frac{1}{2}.$$

Cursory examination of the second term on the RHS of (1) above indicates that $~s~$ must be $~> 0.~$ So, it only remains to sanity check $~s = \dfrac{1}{2}, ~$ using the second term on the RHS of (1) above:

$$\left[ ~\frac{3 \times 9}{4} \times \frac{1}{2} ~\right] + \frac{13}{8} = \frac{27 + 13}{8} = 5.$$

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$\underline{\text{Method 2}}$

Use Quanto's answer to this MathSE denesting question, which is shown below for reference:

Apply the known denesting formula $$\sqrt[3]{a+b \sqrt c}=\frac12\sqrt[3]{3bt-a}\left(1+\frac1t \sqrt c\right)$$

where $t$ satisfies $$t^3-\frac{3a}bt^2+3c t-\frac{ac}b=0. \tag3 $$

Since the expression to denest is $\displaystyle ~\sqrt[3]{18 + 5 \sqrt{13}}, ~$
you have that $~(a,b,c) = (18,5,13).$

Therefore, the cubic to be satisfied, which is represented by (3) above, is actually

$$t^3 + t^2\left[ ~-\frac{54}{5} ~\right] + t[ ~39 ~] + \left[ ~-\frac{234}{5} ~\right] = 0. \implies $$

$$f(t) = t^3[ ~5 ~] + t^2\left[ ~-54 ~\right] + t[ ~195 ~] + \left[ ~-234 ~\right] = 0. \tag4 $$

Note that it is sufficient to find any of the 3 roots to (4) above [i.e. $~f(t) = 0$], to serve as the value of $~t.~$ Naturally, it is preferred to find a real root. In general, there are two methods of solving a cubic equation, such as the one in (4) above:

• Hope that at least one of the roots is rational, and use the rational root theorem.
  This should be your first try.

• Use the method discussed in this article on cubic equations. This should be your last resort.

Using the rational root theorem, you are looking for a satisfying value of $~\displaystyle t = \frac{p}{q},~$ where $~p,q~$ are relatively prime integers, and where:

• $p~$ is a factor of $~234 = 2 \times 3^2 \times 13.$

• $q~$ is a factor of $~5.$

Before engaging in blind guesswork, note that
$f'(t) = 15t^2 - 108t + 195 = 3(5t^2 - 36t + 65)$

$\displaystyle = \frac{3}{5} (25t^2 - 180t + 325) = \frac{3}{5} \left[ ~(5t - 18)^2 + 1 ~\right].~$

So, $~f'(t)~$ is always positive. Therefore, (4) above will have exactly one real root. Also, since $~f(0) = -234, ~$ the root must be positive. Further, since $~f(1) = -88 ~$ and $~f(5) = 16,~$ you must have that $~1 < t < 5.$

Using these results, you can use trial and error on the viable values of $~(p,q)~$ to determine that $~f(3) = 0.~$

Therefore, $~\displaystyle ~\sqrt[3]{18 + 5 \sqrt{13}}~$ is denested to

$$\frac{1}{2} \sqrt[3]{45-18} \left[ ~1 + \frac{1}{3}\sqrt{13} ~\right] = \frac{3}{2} + \frac{1}{2}\sqrt{13}.$$

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$\underline{\text{Addendum-1}}$

In general, I advise sticking with either of the two methods shown above, and I advise against trying to directly use that

$$\left[ ~r + s\sqrt{w} ~\right]^3 = \left[r^3 + (3 rs^2w) \right] + \sqrt{w}\left[ ~3r^2s + ws^3 ~\right]. $$

$$\left[ ~r - s\sqrt{w} ~\right]^3 = \left[r^3 + (3 rs^2w) \right] - \sqrt{w}\left[ ~3r^2s + ws^3 ~\right]. $$

The point is that (to the best of my knowledge), there is no elegant analytical method of solving the simultaneous equations:

$$r^3 + (3 rs^2w) = M, ~~~(3r^2s) + ws^3 = N \\ \text{where} ~~M,N,w ~~\text{are known}.$$

Further, in those situations, where a viable guess to the above equations proves accurate, I suspect (without knowing for sure) that method 2 above will yield a rational root for $~t.~$

Since applying the rational root theorem to a cubic equation is not that bad, and takes very little creativity (unlike the enlightened guesswork approach of attacking two simultaneous non-linear equations), I give preference to method 2.

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$\underline{\text{Addendum-2}}$

A very reasonable perspective is to regard method 1 as off limits, since it uses a calculator. In fact, there is a hybrid approach of method 1 and method 2 that does not involve a calculator, and is arguably superior to using only method 2.

• Examining method 2, you have that
  
  when $~\displaystyle \sqrt[3]{a + b\sqrt{c}}~$ denests to $~\displaystyle r + s\sqrt{c},~$
  
  that $\displaystyle ~r = \frac12\sqrt[3]{3bt-a}.$

• Often, when examining the two terms:
  
  $~\displaystyle \sqrt[3]{a + b\sqrt{c}}~~$ and $~~\displaystyle \sqrt[3]{a - b\sqrt{c}},~~$
  
  one of the two terms will be relatively small.
  
  Further, it will often be fairly easy to mentally approximate (without any calculator) the other term.

So, you will have a rough estimate of

$$r = \frac{1}{2} \sqrt[3]{3bt-a}.$$

In the above equation, with $~a,b~$ known, and $~r~$ approximated, you can approximate $~t.~$

Such an approximation will often streamline the use of the rational root theorem, because it will drastically affect which values of $~(p,q) : t = \dfrac{p}{q}~$ are viable.