If : `f(x+1/x)=x^3+1/x^3,` where `x ne0` then: `f(x)=...`

Options

A. `x^3-3x`
B. `3x^3-x`
C. `3x-x^3`
D. `1-3x^2`

Answer

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

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

where \( x \neq 0 \).

### Step 1: Use the identity for \( a^3 + b^3 \)

Recall the identity for the sum of cubes:

\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]

In our case, let \( a = x \) and \( b = \frac{1}{x} \). Thus, we can express \( x^3 + \frac{1}{x^3} \) as:

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

### Step 2: Simplify the expression

Now simplify the expression inside the parentheses:

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

Thus, we can rewrite \( x^3 + \frac{1}{x^3} \) as:

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

### Step 3: Express \( x^2 + \frac{1}{x^2} \)

We also know that:

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

Let \( t = x + \frac{1}{x} \). Then:

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

### Step 4: Substitute back into the equation

Now substituting this back into our expression for \( x^3 + \frac{1}{x^3} \):

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

### Step 5: Relate back to \( f \)

Thus, we have:

\[ f(t) = t(t^2 - 3) \]

Substituting \( t = x + \frac{1}{x} \), we find:

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

### Step 6: Find \( f(x) \)

To find \( f(x) \), we replace \( t \) with \( x \):

\[ f(x) = x(x^2 - 3) \]

Thus, the final answer is:

\[ f(x) = x^3 - 3x \]

### Final Answer

\[ f(x) = x^3 - 3x \]

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