If ` x = k ^(3) - 3 k^( 2) and y = 1 - 3 k` then for what value of k, will be x = y ?

Options

A. (A) 0
B. (B)1
C. (C) -1`
D. (D)2

Answer

To solve the problem where we need to find the value of \( k \) such that \( x = y \) given the equations \( x = k^3 - 3k^2 \) and \( y = 1 - 3k \), we can follow these steps:

### Step 1: Set the equations equal to each other
We start by equating \( x \) and \( y \):
\[ k^3 - 3k^2 = 1 - 3k \]

### Step 2: Rearrange the equation
Next, we rearrange the equation to bring all terms to one side:
\[ k^3 - 3k^2 + 3k - 1 = 0 \]

### Step 3: Factor the polynomial
Now we need to factor the cubic polynomial \( k^3 - 3k^2 + 3k - 1 \). We can try to find rational roots using the Rational Root Theorem. Testing \( k = 1 \):
\[ 1^3 - 3(1^2) + 3(1) - 1 = 1 - 3 + 3 - 1 = 0 \]
Since \( k = 1 \) is a root, we can factor the polynomial as \( (k - 1)(k^2 - 2k + 1) \).

### Step 4: Factor further
The quadratic \( k^2 - 2k + 1 \) can be factored as:
\[ (k - 1)^2 \]
So, the complete factorization of the polynomial is:
\[ (k - 1)^3 = 0 \]

### Step 5: Solve for \( k \)
Setting the factored equation to zero gives:
\[ k - 1 = 0 \implies k = 1 \]

### Conclusion
Thus, the value of \( k \) for which \( x = y \) is:
\[ \boxed{1} \]