To expand the expression $(2ax - by)^3$, we can use the Binomial Theorem, which states:

$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

For $(2ax - by)^3$, let’s identify the terms:

• $a = 2ax$
• $b = -by$
• $n = 3$

Now, we can apply the Binomial Theorem:

$(2ax - by)^3 = \sum_{k=0}^{3} \binom{3}{k} (2ax)^{3-k} (-by)^k$

Step-by-Step Expansion

1. Term with $k = 0$: $\binom{3}{0} (2ax)^3 (-by)^0 = 1 \cdot (2ax)^3 \cdot 1 = 8a^3x^3$

2. Term with $k = 1$: $\binom{3}{1} (2ax)^2 (-by)^1 = 3 \cdot (4a^2x^2) \cdot (-by) = -12a^2xbxy$

3. Term with $k = 2$: $\binom{3}{2} (2ax)^1 (-by)^2 = 3 \cdot (2ax) \cdot (b^2y^2) = 6ab^2xy^2$

4. Term with $k = 3$: $\binom{3}{3} (2ax)^0 (-by)^3 = 1 \cdot 1 \cdot (-b^3y^3) = -b^3y^3$

Putting It All Together

Combine all the terms:

$(2ax - by)^3 = 8a^3x^3 - 12a^2xbxy + 6ab^2xy^2 - b^3y^3$

Numerical Coefficients

Now, let's identify the numerical coefficients for each term:

1. Blank 1 (for $a^3x^3$): $8$
2. Blank 2 (for $a^2xbxy$): $-12$
3. Blank 3 (for $ab^2xy^2$): $6$
4. Blank 4 (for $b^3y^3$): $-1$

Thus, the numerical coefficients are:

• Blank 1: 8
• Blank 2: $-12$
• Blank 3: 6
• Blank 4: $-1$

Would you like more details or have any questions?

Related Questions:

1. What is the Binomial Theorem, and how is it applied in algebra?
2. Can you explain how to compute binomial coefficients $\binom{n}{k}$?
3. How do signs change in binomial expansions involving subtraction?
4. What are some real-world applications of binomial expansions?
5. How would you expand a binomial raised to the fourth power?

Tip: When expanding expressions using the Binomial Theorem, carefully handle the signs of the terms, especially if they involve subtraction or negative values.