If ` x^(4) + ((1)/(x^(4))) = 34` , what is the value of `x^(3) - ((1)/(x^(3))) ` ?

Options

A. 0
B. 6
C. 8
D. 14

Answer

To solve the problem, we start with the equation given:

\[ x^4 + \frac{1}{x^4} = 34 \]

### Step 1: Rewrite the equation using squares
We can express \( x^4 + \frac{1}{x^4} \) in terms of \( x^2 + \frac{1}{x^2} \). We know that:

\[ x^4 + \frac{1}{x^4} = \left( x^2 + \frac{1}{x^2} \right)^2 - 2 \]

So, we can set up the equation:

\[ \left( x^2 + \frac{1}{x^2} \right)^2 - 2 = 34 \]

### Step 2: Solve for \( x^2 + \frac{1}{x^2} \)
Adding 2 to both sides gives:

\[ \left( x^2 + \frac{1}{x^2} \right)^2 = 36 \]

Taking the square root of both sides, we find:

\[ x^2 + \frac{1}{x^2} = 6 \quad \text{or} \quad x^2 + \frac{1}{x^2} = -6 \quad \text{(not possible since both terms are positive)} \]

Thus, we have:

\[ x^2 + \frac{1}{x^2} = 6 \]

### Step 3: Find \( x - \frac{1}{x} \)
Next, we can express \( x^2 + \frac{1}{x^2} \) in terms of \( x - \frac{1}{x} \):

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

Let \( y = x - \frac{1}{x} \). Then we have:

\[ y^2 + 2 = 6 \]

Subtracting 2 from both sides gives:

\[ y^2 = 4 \]

Taking the square root of both sides, we find:

\[ y = 2 \quad \text{or} \quad y = -2 \]

Since we are looking for \( x - \frac{1}{x} \), we can take:

\[ x - \frac{1}{x} = 2 \]

### Step 4: Find \( x^3 - \frac{1}{x^3} \)
Now, we can use the identity for cubes:

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

We already know \( x - \frac{1}{x} = 2 \) and we can find \( x^2 + 1 + \frac{1}{x^2} \):

\[ x^2 + \frac{1}{x^2} = 6 \implies x^2 + 1 + \frac{1}{x^2} = 6 + 1 = 7 \]

Now substituting back:

\[ x^3 - \frac{1}{x^3} = 2 \cdot 7 = 14 \]

### Final Answer
Thus, the value of \( x^3 - \frac{1}{x^3} \) is:

\[ \boxed{14} \]