4 x y

Answer

To solve the question step by step, we will find the product of the given algebraic expressions.

### Step 1: Identify the expressions to be multiplied
We have two parts to the question. The first part is multiplying \( 6xy \) with \( -3x^3y^3 \).

### Step 2: Multiply the constants
First, we multiply the constants:
\[ 6 \times (-3) = -18 \]

### Step 3: Multiply the variables
Next, we multiply the variables:
- For \( x \):
\[ x^1 \times x^3 = x^{1+3} = x^4 \]
- For \( y \):
\[ y^1 \times y^3 = y^{1+3} = y^4 \]

### Step 4: Combine the results
Now, we combine the results from the constants and variables:
\[ -18x^4y^4 \]

### Answer for the first part:
The answer for the first part is:
\[ -18x^4y^4 \]

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### Step 5: Move to the second part
Now, we will multiply the three algebraic expressions: \( 7ab^2 \), \( -4a^2b \), and \( -5abc \).

### Step 6: Multiply the first two expressions
First, we multiply \( 7ab^2 \) and \( -4a^2b \):
\[ 7 \times (-4) = -28 \]
For the variables:
- For \( a \):
\[ a^1 \times a^2 = a^{1+2} = a^3 \]
- For \( b \):
\[ b^2 \times b^1 = b^{2+1} = b^3 \]

So, the result of multiplying the first two expressions is:
\[ -28a^3b^3 \]

### Step 7: Multiply the result with the third expression
Now, we multiply \( -28a^3b^3 \) with \( -5abc \):
\[ -28 \times (-5) = 140 \]
For the variables:
- For \( a \):
\[ a^3 \times a^1 = a^{3+1} = a^4 \]
- For \( b \):
\[ b^3 \times b^1 = b^{3+1} = b^4 \]
- For \( c \):
\[ c^0 \times c^1 = c^{0+1} = c^1 \]

### Step 8: Combine the results
Now, we combine the results:
\[ 140a^4b^4c \]

### Answer for the second part:
The answer for the second part is:
\[ 140a^4b^4c \]

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### Final Answers:
1. For the first part: \(-18x^4y^4\)
2. For the second part: \(140a^4b^4c\)

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