Question

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

Answer 1

`3(\root(3){3}+\frac{1}{\root(3){3}})^3-9(\root(3){3}+\frac{1}{\root(3){3}})=10`

Explanation:

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

`3^{\frac{1}{3}}=\root(3){3}`

`3^{-\frac{1}{3}}=\frac{1}{3^{\frac{1}{3}}}=\frac{1}{\root(3){3}}`

`\therefore x=(\root(3){3}+\frac{1}{\root(3){3}})`

Plugging in that value for `x` yields:

`3(\root(3){3}+\frac{1}{\root(3){3}})^3-9(\root(3){3}+\frac{1}{\root(3){3}})=10`

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

Answer 2

`0`

Explanation:

`3x^3-9x=10`
Where `x=3^(1/3)+3^(-1/3)`
Of course it's​ answer will be zero if the equation has a solution
Then
`x^3=[3^(1/3)+3^(-1/3)]^3`
`=>[3^(1/3)]^3+[3^(-1/3)]^3+3.3^(1/3).3^(-2/3)+3.3^(2/3).3^(-1/3)`
`=>3^1+3^(-1)+3.3^(-1/3)+3.3^(1/3)`
`=>3+1/3+3(3^(1/3)+3^(-1/3))`
`=>3+1/3+3x`
Substitute the value of `x^3`in given equation
`=>3(3+1/3+3x)-9x=10`
`=>9+1+9x-9x=10`
`10=10`
`10-10=0`