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

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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