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

2 Answers
Morgan S.
Oct 9, 2017

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.

aniket r.
Oct 9, 2017

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^3in given equation
=>3(3+1/3+3x)-9x=10
=>9+1+9x-9x=10
10=10
10-10=0

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