To expand $5(x^2 - 2y^3)^3$, we will first apply the binomial expansion formula to the expression $(x^2 - 2y^3)^3$ and then multiply by 5.

The binomial expansion formula for $(a - b)^n$ is:

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

In this case, $a = x^2$, $b = 2y^3$, and $n = 3$.

Step 1: Apply the binomial expansion

$(x^2 - 2y^3)^3 = \sum_{k=0}^{3} \binom{3}{k} (x^2)^{3-k} (-2y^3)^k$

Now, we compute each term:

• For $k=0$: $\binom{3}{0} (x^2)^{3-0} (-2y^3)^0 = 1 \cdot x^6 \cdot 1 = x^6$

• For $k=1$: $\binom{3}{1} (x^2)^{3-1} (-2y^3)^1 = 3 \cdot x^4 \cdot (-2y^3) = -6x^4 y^3$

• For $k=2$: $\binom{3}{2} (x^2)^{3-2} (-2y^3)^2 = 3 \cdot x^2 \cdot 4y^6 = 12x^2 y^6$

• For $k=3$: $\binom{3}{3} (x^2)^{3-3} (-2y^3)^3 = 1 \cdot 1 \cdot (-8y^9) = -8y^9$

Step 2: Combine the terms

$(x^2 - 2y^3)^3 = x^6 - 6x^4 y^3 + 12x^2 y^6 - 8y^9$

Step 3: Multiply by 5

Now, we multiply each term by 5:

$5(x^2 - 2y^3)^3 = 5(x^6 - 6x^4 y^3 + 12x^2 y^6 - 8y^9)$

$= 5x^6 - 30x^4 y^3 + 60x^2 y^6 - 40y^9$

So, the expanded form of $5(x^2 - 2y^3)^3$ is:

$\boxed{5x^6 - 30x^4 y^3 + 60x^2 y^6 - 40y^9}$

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Would you like further details or clarification on any of the steps?

Here are some related questions you can explore:

1. How can we use the binomial theorem for other powers, such as $(x + y)^4$?
2. What would the expansion of $(x^2 + 3y^3)^4$ look like?
3. How do we handle higher powers of binomials in multivariable calculus?
4. Can we apply the binomial expansion to non-integer exponents?
5. What are the applications of binomial expansions in physics or engineering?

Tip: Practice applying the binomial theorem to different powers to get more comfortable with the pattern and coefficients!