If `x=root3(a)-1/(root3(a))` then value of `x^3 +3x` is

Options

A. `a+a^(-1)`
B. `a-a^(-1)`
C. `a^2+a^(-2)`
D. 0

Answer

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

\[ x = \sqrt[3]{a} - \frac{1}{\sqrt[3]{a}} \]

We need to find the value of \( x^3 + 3x \).

### Step 1: Calculate \( x^3 \)

Using the formula for the cube of a binomial, we have:

\[ x^3 = \left( \sqrt[3]{a} - \frac{1}{\sqrt[3]{a}} \right)^3 \]

Using the identity \( (p - q)^3 = p^3 - 3p^2q + 3pq^2 - q^3 \), where \( p = \sqrt[3]{a} \) and \( q = \frac{1}{\sqrt[3]{a}} \):

\[ x^3 = \left(\sqrt[3]{a}\right)^3 - 3\left(\sqrt[3]{a}\right)^2\left(\frac{1}{\sqrt[3]{a}}\right) + 3\left(\sqrt[3]{a}\right)\left(\frac{1}{\sqrt[3]{a}}\right)^2 - \left(\frac{1}{\sqrt[3]{a}}\right)^3 \]

Calculating each term:

1. \( \left(\sqrt[3]{a}\right)^3 = a \)
2. \( 3\left(\sqrt[3]{a}\right)^2\left(\frac{1}{\sqrt[3]{a}}\right) = 3\sqrt[3]{a^2} \cdot \frac{1}{\sqrt[3]{a}} = 3\sqrt[3]{a} \)
3. \( 3\left(\sqrt[3]{a}\right)\left(\frac{1}{\sqrt[3]{a}}\right)^2 = 3\sqrt[3]{a} \cdot \frac{1}{\sqrt[3]{a^2}} = 3\frac{1}{\sqrt[3]{a}} \)
4. \( \left(\frac{1}{\sqrt[3]{a}}\right)^3 = \frac{1}{a} \)

Putting it all together:

\[ x^3 = a - 3\sqrt[3]{a} + 3\frac{1}{\sqrt[3]{a}} - \frac{1}{a} \]

### Step 2: Combine \( x^3 \) and \( 3x \)

Now, we need to add \( 3x \) to \( x^3 \):

\[ 3x = 3\left(\sqrt[3]{a} - \frac{1}{\sqrt[3]{a}}\right) = 3\sqrt[3]{a} - 3\frac{1}{\sqrt[3]{a}} \]

Now, combine \( x^3 \) and \( 3x \):

\[ x^3 + 3x = \left(a - 3\sqrt[3]{a} + 3\frac{1}{\sqrt[3]{a}} - \frac{1}{a}\right) + \left(3\sqrt[3]{a} - 3\frac{1}{\sqrt[3]{a}}\right) \]

Simplifying this:

\[ x^3 + 3x = a - \frac{1}{a} \]

### Final Result

Thus, the value of \( x^3 + 3x \) is:

\[ \boxed{a - \frac{1}{a}} \]