# Show that $\sqrt{8-4\sqrt3} =\sqrt[6]2\:\! {\sqrt[3]{12\sqrt3-20}}$

This was the result of evaluating an integral by two different methods. The RHS was obtained by making a substitution, the LHS was obtained using trigonometric identity's and partial fractions.

Now I know that these two are equal, but I just can't prove it. I tried writing the LHS in terms of powers of $2$, but can't get any further to the desired result.

• What was the integral ? Jun 22, 2014 at 20:29
• Seems unlikely, but they are equal: wolfram
– user122283
Jun 22, 2014 at 20:30
• A slight simplification reduces the quantity to $$\frac{\sqrt{2-\sqrt 3}}{\sqrt[3]{3\sqrt3-5}}=2^{-1\over 6}$$ Jun 22, 2014 at 20:32
• Yes thats exactly what i obtained but cannot get any further ----thanks Jun 22, 2014 at 20:36

We have

$$8 - 4\sqrt{3} = 2(4-2\sqrt{3}) = 2(3-2\sqrt{3}+1) = 2(\sqrt{3}-1)^2,$$

so $\sqrt{8-4\sqrt{3}} = \sqrt{2}(\sqrt{3}-1)$.

Then we see that looking at $(\sqrt{3}-1)^3$ is a good idea:

$$(\sqrt{3}-1)^3 = (4-2\sqrt{3})(\sqrt{3}-1) = 6\sqrt{3} - 10,$$

so $12\sqrt{3}-20 = 2(\sqrt{3}-1)^3$ and $\sqrt[3]{12\sqrt{3}-20} = \sqrt[3]{2}(\sqrt{3}-1)$.

$\frac{\sqrt{2}}{\sqrt[3]{2}} = 2^{1/6}$ is then easy to see.

• Very impressive, don't know how you answered that so quickly, thank you. Jun 22, 2014 at 20:42
• So the question remains is that is it possible to always be able to prove that one expression is equal to another when we know that they are? Jun 22, 2014 at 20:48
• That depends on what you mean by "know". In a strict sense, you only know that two expressions are equal if you have a proof of that fact. If you take "know" less strictly, I suppose there are expressions one "knows" to be equal without being able to prove it. But for expressions of the kind involved here, involving only roots of rational numbers and arithmetic operations, in principle one can (I think) always prove equality, though sometimes that can be really difficult. Jun 22, 2014 at 20:53

$$\frac{\sqrt{8-4\sqrt3}}{\sqrt[3]{12\sqrt3-20}} =\sqrt[6]{\frac{4^3(2-\sqrt3)^3}{4^2(3\sqrt3-5)^2}}=\sqrt[6]{\frac{4(2-\sqrt3)^3}{(3\sqrt3-5)^2}}=\sqrt[6]{\frac{4(8-12\sqrt3+18-3\sqrt3)}{(27-30\sqrt3+25)}}=$$ $$\sqrt[6]{\frac{4(26-15\sqrt3)}{(52-30\sqrt3)}}=\sqrt[6]{\frac{4(26-15\sqrt3)}{2(26-15\sqrt3)}}=2^{\frac{1}{6}}$$

• Excellent answer Jun 22, 2014 at 20:44
• Thank you, so the question remains is that is it possible to always be able to prove that one expression is equal to another when we know that they are? Jun 22, 2014 at 20:47
• @user144533 yes, such algorithms do exist - see my answer. Jun 23, 2014 at 0:50

Hint  It arises by taking a $$6$$'th root $$\,(x^{\large \color{#0a0}2})^{\large \color{#c00}3} =\:\! (x^{\large \color{#c00}3})^{\large \color{#0a0}2},\$$ i.e.

\begin{align}(8 - 4\sqrt 3)^{\large \color{#c00}3} =\, ((\sqrt 6 - \sqrt 2)^{\large \color{#0a0}2})^{\large \color{#c00}3} &=\, ((\sqrt 6 - \sqrt 2)^{\large \color{#c00}3})^{\large \color{#0a0}2} =\, 2\, (12\sqrt 3 - 20)^{\large \color{#0a0}2}\\[1em] {\rm so}\ \ \ {\small\rm LHS}^{1/6}\! = {\small\rm RHS}^{1/6}\Rightarrow \sqrt{8-4\sqrt 3} &= \sqrt[6]2\sqrt[3]{12\sqrt{3}-20} \end{align}\qquad\qquad\ \

Remark  This shows how to discover such equalities (vs. verify them by taking $$6$$'th powers).

As to your general question, yes, there are effective algorithms for deciding equality of real algebraic numbers. But they would not be easy to describe at the algebra-precalculus level. For one such algorithm see Renaud Rioboo, Towards faster real algebraic numbers.

Take the sixth power of $$\,x = \sqrt{8-4\sqrt3}\,/\:\!\sqrt[3]{12\sqrt3-20}\,$$ then clear the outside surds $$x^6=\frac {4^3(2-\sqrt 3)^3}{4^2(3\sqrt 3-5)^2}=4\cdot\frac {26-15\sqrt 3}{52-30\sqrt 3}=2$$

The anticipated next step would be to rationalise the denominator, but fortunately the answer drops out easily from the computation.

To address the more general points at the end.

If we know two expressions are equal what does that mean, but that we have a method of proof? So if we know two expressions are equal because we approximate them using a calculator, we can use the method of approximation to create sequences convergent to the same limit.

Does my above "proof" work - well it depends on the fact that our original expression is a positive real number - somehow we have to pick out which of the sixth roots of $$2$$ we mean, so we have to do more work than is initially apparent, and which may bring in issues which are separate from the "purely algebraic" mode of proof - completeness, for the existence of the value and aspects of order too.

${\sqrt{8-4\sqrt3}\over\sqrt[3]{-20+12\sqrt3}}$ =${\left(8-4\sqrt3\right)^{1\over2}\over\left(-20+12\sqrt3\right)^{1\over3}}$ =${\left(8-4\sqrt3\right)^{3\over6}\over\left(-20+12\sqrt3\right)^{2\over6}}$ =$\left(\left(8-4\sqrt3\right)^3\over\left(-20+12\sqrt3\right)^2\right)^\frac16$ =$\sqrt[6]{\left(8-4\sqrt3\right)^3\over\left(-20+12\sqrt3\right)^2}$ =$\sqrt[6]{{1664-960\sqrt3}\over{832-480\sqrt3}}$ =$\sqrt[6]2$