If `x=3^((1)/(3))-3^(-(1)/(3))`, then `(3x^(3)+9x)` is equal to

Options

A. 5
B. 6
C. 7
D. 8

Answer

To solve the problem step by step, we start with the given expression for \( x \):

### Step 1: Define \( x \)
Given:
\[ x = 3^{\frac{1}{3}} - 3^{-\frac{1}{3}} \]

### Step 2: Rewrite \( x \)
We can express \( x \) in terms of a single base:
\[ x = 3^{\frac{1}{3}} - \frac{1}{3^{\frac{1}{3}}} = 3^{\frac{1}{3}} - \frac{1}{3^{\frac{1}{3}}} = 3^{\frac{1}{3}} - 3^{-\frac{1}{3}} \]

### Step 3: Calculate \( x^3 \)
To find \( 3x^3 + 9x \), we first need to calculate \( x^3 \):
Using the identity \( (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \):
Let \( a = 3^{\frac{1}{3}} \) and \( b = 3^{-\frac{1}{3}} \):
\[ x^3 = \left(3^{\frac{1}{3}} - 3^{-\frac{1}{3}}\right)^3 = (3^{\frac{1}{3}})^3 - 3(3^{\frac{1}{3}})^2(3^{-\frac{1}{3}}) + 3(3^{\frac{1}{3}})(3^{-\frac{1}{3}})^2 - (3^{-\frac{1}{3}})^3 \]
Calculating each term:
\[ = 3 - 3 \cdot 3^{\frac{2}{3}} \cdot 3^{-\frac{1}{3}} + 3 \cdot 3^{\frac{1}{3}} \cdot 3^{-\frac{2}{3}} - 3^{-1} \]
\[ = 3 - 3 \cdot 3^{\frac{1}{3}} + 3 \cdot 3^{-\frac{1}{3}} - \frac{1}{3} \]
\[ = 3 - 3x - \frac{1}{3} \]
Combining terms:
\[ = \frac{9}{3} - \frac{1}{3} - 3x = \frac{8}{3} - 3x \]

### Step 4: Substitute \( x^3 \) into \( 3x^3 + 9x \)
Now substitute \( x^3 \) into the expression \( 3x^3 + 9x \):
\[ 3x^3 + 9x = 3\left(\frac{8}{3} - 3x\right) + 9x \]
Distributing:
\[ = 8 - 9x + 9x = 8 \]

### Final Answer
Thus, the value of \( 3x^3 + 9x \) is:
\[ \boxed{8} \]