Find the polynomials `u(x) ` and `v(x)` such that `(x^(4) -1) * u(x) + (x^(7) -1) * v(x) = (x-1)`.

Answer

To find the polynomials \( u(x) \) and \( v(x) \) such that

\[ (x^4 - 1) u(x) + (x^7 - 1) v(x) = (x - 1), \]

we will follow a systematic approach.

### Step 1: Rewrite the equation
We start with the equation:

\[ (x^4 - 1) u(x) + (x^7 - 1) v(x) = (x - 1). \]

### Step 2: Factor the polynomials
Notice that \( x^4 - 1 \) can be factored as:

\[ x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1). \]

Similarly, \( x^7 - 1 \) can be factored using the formula for the difference of powers:

\[ x^7 - 1 = (x - 1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1). \]

### Step 3: Substitute and simplify
Substituting these factorizations into the equation gives:

\[ ((x - 1)(x + 1)(x^2 + 1)) u(x) + ((x - 1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)) v(x) = (x - 1). \]

### Step 4: Factor out \( (x - 1) \)
We can factor \( (x - 1) \) out from both sides:

\[ (x - 1) \left( (x + 1)(x^2 + 1) u(x) + (x^6 + x^5 + x^4 + x^3 + x^2 + x + 1) v(x) \right) = (x - 1). \]

### Step 5: Divide by \( (x - 1) \)
Assuming \( x \neq 1 \), we can divide both sides by \( (x - 1) \):

\[ (x + 1)(x^2 + 1) u(x) + (x^6 + x^5 + x^4 + x^3 + x^2 + x + 1) v(x) = 1. \]

### Step 6: Choose \( u(x) \) and \( v(x) \)
To solve this equation, we can choose \( u(x) \) and \( v(x) \) such that they are simple polynomials. Let's try:

1. Let \( u(x) = x^4 + 1 \).
2. Let \( v(x) = -x \).

### Step 7: Verify the solution
Now we substitute \( u(x) \) and \( v(x) \) back into the equation:

\[ (x^4 - 1)(x^4 + 1) + (x^7 - 1)(-x). \]

Calculating each term:

1. \( (x^4 - 1)(x^4 + 1) = x^8 - 1 \).
2. \( (x^7 - 1)(-x) = -x^8 + x \).

Combining these gives:

\[ x^8 - 1 - x^8 + x = x - 1. \]

This confirms that our choice of \( u(x) \) and \( v(x) \) is correct.

### Final Answer
Thus, the polynomials are:

\[ u(x) = x^4 + 1, \quad v(x) = -x. \]