Factorise : (i) `y^3 - 3y^2 + 2y - 6 - xy + 3x`

Answer

To factorise the expression \( y^3 - 3y^2 + 2y - 6 - xy + 3x \), we can follow these steps:

### Step 1: Rearrange the expression
First, we can rearrange the terms in the expression for better visibility:
\[ y^3 - 3y^2 + 2y - xy + 3x - 6 \]

### Step 2: Group the terms
Next, we can group the terms in a way that allows us to factor them:
\[ (y^3 - 3y^2) + (2y - 6) + (-xy + 3x) \]

### Step 3: Factor out common terms from each group
Now, we will factor out the common terms from each group:
1. From \( y^3 - 3y^2 \), we can factor out \( y^2 \):
\[ y^2(y - 3) \]
2. From \( 2y - 6 \), we can factor out \( 2 \):
\[ 2(y - 3) \]
3. From \( -xy + 3x \), we can factor out \( -x \):
\[ -x(y - 3) \]

Putting it all together, we have:
\[ y^2(y - 3) + 2(y - 3) - x(y - 3) \]

### Step 4: Factor out the common binomial factor
Now we can see that \( (y - 3) \) is a common factor:
\[ (y - 3)(y^2 + 2 - x) \]

### Final Result
Thus, the factorised form of the expression \( y^3 - 3y^2 + 2y - 6 - xy + 3x \) is:
\[ (y - 3)(y^2 + 2 - x) \]