Given that `(1-x) (1+x+x^(2) +x^(3) +x^(4)) = 31/32` and x is a rational number. Then `1+x+x^(2) + x^(3) + x^(4)+ x^(5)` is:

Answer

To solve the equation \( (1-x)(1+x+x^2+x^3+x^4) = \frac{31}{32} \) and find the value of \( 1+x+x^2+x^3+x^4+x^5 \), we can follow these steps:

### Step 1: Simplify the given equation
We start with the equation:
\[ (1-x)(1+x+x^2+x^3+x^4) = \frac{31}{32} \]
We can expand the left-hand side:
\[ 1+x+x^2+x^3+x^4 - x(1+x+x^2+x^3+x^4) = \frac{31}{32} \]
This simplifies to:
\[ 1+x+x^2+x^3+x^4 - (x + x^2 + x^3 + x^4 + x^5) = \frac{31}{32} \]
Now, combining like terms gives:
\[ 1 - x^5 = \frac{31}{32} \]

### Step 2: Isolate \( x^5 \)
Rearranging the equation, we have:
\[ 1 - \frac{31}{32} = x^5 \]
Calculating the left side:
\[ 1 = \frac{32}{32} \quad \Rightarrow \quad \frac{32}{32} - \frac{31}{32} = x^5 \]
Thus:
\[ x^5 = \frac{1}{32} \]

### Step 3: Solve for \( x \)
To find \( x \), we take the fifth root of both sides:
\[ x = \left(\frac{1}{32}\right)^{\frac{1}{5}} = \frac{1}{2} \]

### Step 4: Calculate \( 1+x+x^2+x^3+x^4+x^5 \)
Now we need to find:
\[ 1 + x + x^2 + x^3 + x^4 + x^5 \]
Substituting \( x = \frac{1}{2} \):
\[ 1 + \frac{1}{2} + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^3 + \left(\frac{1}{2}\right)^4 + \left(\frac{1}{2}\right)^5 \]
Calculating each term:
\[ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} \]

### Step 5: Find a common denominator
The common denominator for these fractions is 32:
\[ = \frac{32}{32} + \frac{16}{32} + \frac{8}{32} + \frac{4}{32} + \frac{2}{32} + \frac{1}{32} \]
Adding these fractions:
\[ = \frac{32 + 16 + 8 + 4 + 2 + 1}{32} = \frac{63}{32} \]

### Final Answer
Thus, the value of \( 1+x+x^2+x^3+x^4+x^5 \) is:
\[ \frac{63}{32} \]