# How do you prove this equality sqrt(3+sqrt(3)+(10+6sqrt(3))^(2/3))=sqrt(3)+1 ?

Jul 11, 2018

See here

#### Explanation:

It doesn't look like they're equal, so you can't prove they are..

Jul 11, 2018

The given equation is false, but we can prove:

$\sqrt{3 + \sqrt{3} + {\left(10 + 6 \sqrt{3}\right)}^{\textcolor{red}{\frac{1}{3}}}} = \sqrt{3} + 1$

#### Explanation:

Given to prove:

$\sqrt{3 + \sqrt{3} + {\left(10 + 6 \sqrt{3}\right)}^{\frac{2}{3}}} = \sqrt{3} + 1$

We can perform a sequence of reversible steps until we reach a simplified equation that is known to be true.

Noting that both sides of the equation to be proved are positive, it is reversible to square both sides to get:

$3 + \sqrt{3} + {\left(10 + 6 \sqrt{3}\right)}^{\frac{2}{3}} = 3 + 2 \sqrt{3} + 1 = 4 + 2 \sqrt{3}$

Subtracting $3 + \sqrt{3}$ from both ends, this becomes:

${\left(10 + 6 \sqrt{3}\right)}^{\frac{2}{3}} = 1 + \sqrt{3}$

Raising both sides to the power $3$, this becomes:

${\left(10 + 6 \sqrt{3}\right)}^{2} = 1 + 3 \sqrt{3} + 9 + 3 \sqrt{3} = 10 + 6 \sqrt{3}$

Note that on the left hand side we have ${\left(10 + 6 \sqrt{3}\right)}^{2}$ while on the right hand side we have $10 + 6 \sqrt{3}$

It seems that the exponent $\frac{2}{3}$ should have been $\frac{1}{3}$.

Then we could write a proof as follows:

$10 + 6 \sqrt{3} = 1 + 3 \sqrt{3} + 9 + 3 \sqrt{3}$

color(white)(10+6sqrt(3)) = 1+3(sqrt(3))+3(sqrt(3)^2+(sqrt(3))^3

$\textcolor{w h i t e}{10 + 6 \sqrt{3}} = {\left(1 + \sqrt{3}\right)}^{3}$

Taking the cube root of both ends, we find:

${\left(10 + 6 \sqrt{3}\right)}^{\frac{1}{3}} = 1 + \sqrt{3}$

Adding $3 + \sqrt{3}$ to both sides, this becomes:

$3 + \sqrt{3} + {\left(10 + 6 \sqrt{3}\right)}^{\frac{1}{3}} = 3 + 2 \sqrt{3} + 1$

$3 + \sqrt{3} + {\left(10 + 6 \sqrt{3}\right)}^{\frac{1}{3}} = {\left(\sqrt{3} + 1\right)}^{2}$

Then taking the square root of both sides, we find:

$\sqrt{3 + \sqrt{3} + {\left(10 + 6 \sqrt{3}\right)}^{\frac{1}{3}}} = \left\mid \sqrt{3} + 1 \right\mid = \sqrt{3} + 1$

Jul 11, 2018

Please see below if it is $\sqrt{3 + \sqrt{3} + {\left(10 + 6 \sqrt{3}\right)}^{\frac{1}{3}}} = \sqrt{3} + 1$

#### Explanation:

It should have been $\sqrt{3 + \sqrt{3} + {\left(10 + 6 \sqrt{3}\right)}^{\frac{1}{3}}} = \sqrt{3} + 1$

Observe that ${\left(1 + \sqrt{3}\right)}^{3} = {1}^{3} + 3 \times {1}^{2} \times \sqrt{3} + 3 \times 1 \times 3 + {\left(\sqrt{3}\right)}^{3}$

= $1 + 3 \sqrt{3} + 9 + 3 \sqrt{3} = 10 + 3 \sqrt{3}$

Hence $\sqrt{3 + \sqrt{3} + {\left(10 + 6 \sqrt{3}\right)}^{\frac{2}{3}}}$

= $\sqrt{3 + \sqrt{3} + 1 + \sqrt{3}}$

= $\sqrt{{\left(\sqrt{3}\right)}^{2} + 2 \times \sqrt{3} \times 1 + {1}^{2}}$

= $\sqrt{{\left(\sqrt{3} + 1\right)}^{2}}$

= $\sqrt{3} + 1$