# If x=3^(1/3)+3^(-1/3) then what is the value of 3x^3-9x=10?

Oct 9, 2017

$3 {\left(\setminus \sqrt{3} + \setminus \frac{1}{\setminus \sqrt{3}}\right)}^{3} - 9 \left(\setminus \sqrt{3} + \setminus \frac{1}{\setminus \sqrt{3}}\right) = 10$

#### Explanation:

Let's solve each term on the RHS of the first equation:

${3}^{\setminus \frac{1}{3}} = \setminus \sqrt{3}$

${3}^{- \setminus \frac{1}{3}} = \setminus \frac{1}{{3}^{\setminus \frac{1}{3}}} = \setminus \frac{1}{\setminus \sqrt{3}}$

$\setminus \therefore x = \left(\setminus \sqrt{3} + \setminus \frac{1}{\setminus \sqrt{3}}\right)$

Plugging in that value for $x$ yields:

$3 {\left(\setminus \sqrt{3} + \setminus \frac{1}{\setminus \sqrt{3}}\right)}^{3} - 9 \left(\setminus \sqrt{3} + \setminus \frac{1}{\setminus \sqrt{3}}\right) = 10$

That's the extent of my knowledge. I'll have someone check and add on to this answer.

Oct 9, 2017

$0$

#### Explanation:

$3 {x}^{3} - 9 x = 10$
Where $x = {3}^{\frac{1}{3}} + {3}^{- \frac{1}{3}}$
Of course it's​ answer will be zero if the equation has a solution
Then
${x}^{3} = {\left[{3}^{\frac{1}{3}} + {3}^{- \frac{1}{3}}\right]}^{3}$
$\implies {\left[{3}^{\frac{1}{3}}\right]}^{3} + {\left[{3}^{- \frac{1}{3}}\right]}^{3} + {3.3}^{\frac{1}{3}} {.3}^{- \frac{2}{3}} + {3.3}^{\frac{2}{3}} {.3}^{- \frac{1}{3}}$
$\implies {3}^{1} + {3}^{- 1} + {3.3}^{- \frac{1}{3}} + {3.3}^{\frac{1}{3}}$
$\implies 3 + \frac{1}{3} + 3 \left({3}^{\frac{1}{3}} + {3}^{- \frac{1}{3}}\right)$
$\implies 3 + \frac{1}{3} + 3 x$
Substitute the value of ${x}^{3}$in given equation
$\implies 3 \left(3 + \frac{1}{3} + 3 x\right) - 9 x = 10$
$\implies 9 + 1 + 9 x - 9 x = 10$
$10 = 10$
$10 - 10 = 0$