The given expression to expand is $(3x^2 - 4y^3)^4$, and the task is to use Pascal's triangle to expand it.

Step-by-Step Expansion Using Pascal's Triangle:

1. Pascal's Triangle provides the binomial coefficients for expansion. For $(a - b)^4$, the row for $n = 4$ in Pascal's Triangle is: $1, 4, 6, 4, 1$

   These are the coefficients for the expansion.

2. Structure of the expansion: The general binomial expansion formula for $(a + b)^n$ is: $(a + b)^n = C_0 a^n + C_1 a^{n-1} b + C_2 a^{n-2} b^2 + \cdots + C_n b^n$ where $C_k$ are the binomial coefficients.

3. Substitute $a = 3x^2$ and $b = -4y^3$ into the formula: $(3x^2 - 4y^3)^4 = 1 \cdot (3x^2)^4 + 4 \cdot (3x^2)^3(-4y^3) + 6 \cdot (3x^2)^2(-4y^3)^2 + 4 \cdot (3x^2)(-4y^3)^3 + 1 \cdot (-4y^3)^4$

4. Calculate each term:

   • First term: $(3x^2)^4 = 81x^8$
   • Second term: $4 \cdot (3x^2)^3(-4y^3) = 4 \cdot 27x^6 \cdot (-4y^3) = -432x^6y^3$
   • Third term: $6 \cdot (3x^2)^2(-4y^3)^2 = 6 \cdot 9x^4 \cdot 16y^6 = 864x^4y^6$
   • Fourth term: $4 \cdot (3x^2)(-4y^3)^3 = 4 \cdot 3x^2 \cdot (-64y^9) = -768x^2y^9$
   • Fifth term: $(-4y^3)^4 = 256y^{12}$

5. Combine all terms: $(3x^2 - 4y^3)^4 = 81x^8 - 432x^6y^3 + 864x^4y^6 - 768x^2y^9 + 256y^{12}$

Thus, the expanded form of $(3x^2 - 4y^3)^4$ is: $81x^8 - 432x^6y^3 + 864x^4y^6 - 768x^2y^9 + 256y^{12}$

Would you like any additional details or have any questions about this expansion?

Related Questions:

1. How does Pascal's triangle apply to binomial expansions of other powers?
2. Can you explain how the signs change when expanding binomials with subtraction?
3. How would the expansion differ for $(3x^2 + 4y^3)^4$ instead?
4. What happens if the exponents in the binomial terms are different (e.g., $3x^2 - 4y$)?
5. How can you generalize this method for any binomial raised to a power?

Tip: Pascal's triangle not only helps with expansion but also provides insights into combinatorial problems.