# How do you simplify root(3)(135+78sqrt(3))+root(3)(135-78sqrt(3)) ?

Apr 20, 2018

$\sqrt{135 + 78 \sqrt{3}} + \sqrt{135 - 78 \sqrt{3}} = 6$

#### Explanation:

Let:

$x = \sqrt{135 + 78 \sqrt{3}} + \sqrt{135 - 78 \sqrt{3}}$

Then:

${x}^{3} = \left(135 + 78 \sqrt{3}\right) + \left(135 - 78 \sqrt{3}\right) + 3 \left(\sqrt{135 + 78 \sqrt{3}} \sqrt{135 - 78 \sqrt{3}}\right) \left(\sqrt{135 + 78 \sqrt{3}} + \sqrt{135 - 78 \sqrt{3}}\right)$

$\textcolor{w h i t e}{{x}^{3}} = 270 + 3 \left(\sqrt{{135}^{2} - {78}^{2} \left(3\right)}\right) x$

$\textcolor{w h i t e}{{x}^{3}} = 270 + 3 \left(\sqrt{18225 - 18252}\right) x$

$\textcolor{w h i t e}{{x}^{3}} = 270 + 3 \left(\sqrt{- 27}\right) x$

$\textcolor{w h i t e}{{x}^{3}} = 270 - 9 x$

So:

${x}^{3} + 9 x - 270 = 0$

By the rational roots theorem, any rational roots of this cubic are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 270$ and $q$ a divisor of the coefficient $1$ of the leading term.

So the first few possible rational zeros are:

$\pm 1 , \pm 2 , \pm 3 , \pm 5 , \pm 6 , \pm 9 , \pm 10 , \ldots$

By Descartes' Rule of Signs, we can tell that the cubic has exactly one positive real root and no negative real zeros, so try the possible positive rational zeros.

We find:

${\left(\textcolor{b l u e}{6}\right)}^{3} + 9 \left(\textcolor{b l u e}{6}\right) - 270 = 216 + 54 - 270 = 0$

So $x = 6$ is a root and $\left(x - 6\right)$ is a factor:

${x}^{3} + 9 x - 270 = \left(x - 6\right) \left({x}^{2} + 6 x + 45\right)$

The remaining quadratic has no real zeros (as we can see by looking at its discriminant).

The non-real zeros are actually:

$\omega \sqrt{135 + 78 \sqrt{3}} + {\omega}^{2} \sqrt{135 - 78 \sqrt{3}}$

and:

${\omega}^{2} \sqrt{135 + 78 \sqrt{3}} + \omega \sqrt{135 - 78 \sqrt{3}}$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive complex cube root of $1$.

So:

$\sqrt{135 + 78 \sqrt{3}} + \sqrt{135 - 78 \sqrt{3}} = 6$

Apr 20, 2018

$6$.

#### Explanation:

Let us note that,

${\left(x + \sqrt{3} y\right)}^{3} = {x}^{3} + 3 \sqrt{3} {y}^{3} + 3 \sqrt{3} x y \left(x + \sqrt{3} y\right)$,

$= {x}^{3} + 3 \sqrt{3} {y}^{3} + 3 \sqrt{3} {x}^{2} y + 9 x {y}^{2}$.

$\Rightarrow {\left(x + \sqrt{3} y\right)}^{3} = \left({x}^{3} + 9 x {y}^{2}\right) + 3 \sqrt{3} \left({x}^{2} y + {y}^{3}\right) , \mathmr{and} ,$

$\sqrt{\left({x}^{3} + 9 x {y}^{2}\right) + \sqrt{3} \left(3 {x}^{2} y + 3 {y}^{3}\right)} = x + \sqrt{3} y$.

So, from this we guess :

If $\sqrt{135 + 78 \sqrt{3}} = x + \sqrt{3} y$, then,

$\left({x}^{3} + 9 x {y}^{2}\right) = 135 , \mathmr{and} , \left(3 {x}^{2} y + 3 {y}^{3}\right) = 78 , i . e . ,$

${x}^{3} + 9 x {y}^{2} = x \left({x}^{2} + 9 {y}^{2}\right) = 135. \ldots . . \left(1\right) , \mathmr{and} ,$

$3 y \left({x}^{2} + {y}^{2}\right) = 78 , \mathmr{and} , y \left({x}^{2} + {y}^{2}\right) = 26. \ldots . . \left(2\right)$.

$y \left({x}^{2} + {y}^{2}\right) = 26 = 1 \times 26 , \mathmr{and} , 2 \times 13$

By Trial and Error Method,  y=1, x=5;" not suitable for "(1)

$y = 2 , x = 3 \text{ satisfy } \left(1\right)$.

Therefore, $\sqrt{135 + 78 \sqrt{3}} = x + \sqrt{3} y = 3 + 2 \sqrt{3}$.

Consequently, $\sqrt{135 - 78 \sqrt{3}} = 3 - 2 \sqrt{3}$.

Finally, $\sqrt{135 + 78 \sqrt{3}} + \sqrt{135 - 78 \sqrt{3}} = 6$,

as Respected George C. Sir has already derived!