When divided by x- 3 the polynomials `x^3-px^2+x+6 and 2x^3-x^2-(p+3)x-6` leave the same remainder. Find the value of 'p'.

Answer

To solve the problem, we need to find the value of \( p \) such that the polynomials \( P(x) = x^3 - px^2 + x + 6 \) and \( G(x) = 2x^3 - x^2 - (p + 3)x - 6 \) leave the same remainder when divided by \( x - 3 \).

### Step 1: Use the Remainder Theorem
According to the Remainder Theorem, the remainder of a polynomial \( f(x) \) when divided by \( x - c \) is \( f(c) \). Here, we will evaluate both polynomials at \( x = 3 \).

### Step 2: Evaluate \( P(3) \)
Substituting \( x = 3 \) into \( P(x) \):
\[ P(3) = 3^3 - p(3^2) + 3 + 6 \]
Calculating each term:
\[ P(3) = 27 - 9p + 3 + 6 = 36 - 9p \]

### Step 3: Evaluate \( G(3) \)
Now, substituting \( x = 3 \) into \( G(x) \):
\[ G(3) = 2(3^3) - (3^2) - (p + 3)(3) - 6 \]
Calculating each term:
\[ G(3) = 2(27) - 9 - (3p + 9) - 6 \]
\[ G(3) = 54 - 9 - 3p - 9 - 6 = 30 - 3p \]

### Step 4: Set the Remainders Equal
Since the remainders are the same, we can set \( P(3) \) equal to \( G(3) \):
\[ 36 - 9p = 30 - 3p \]

### Step 5: Solve for \( p \)
Rearranging the equation:
\[ 36 - 30 = -3p + 9p \]
\[ 6 = 6p \]
Dividing both sides by 6:
\[ p = 1 \]

### Final Answer
Thus, the value of \( p \) is \( \boxed{1} \).