To solve the problem, we need to expand the given expression $(1 + 2x - 2x^2)^8$ using the Binomial Theorem, stopping at the term involving $x^3$.

Step 1: Recall the binomial theorem

The Binomial Theorem states: $(a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$ Here, $a = 1$, $b = 2x - 2x^2$, and $n = 8$.

Step 2: Expand using the binomial theorem

We focus on expanding terms up to $x^3$, combining powers of $x$ arising from $b^k = (2x - 2x^2)^k$.

First term ($k = 0$):

$\binom{8}{0} (1)^{8} (2x - 2x^2)^0 = 1$

Second term ($k = 1$):

$\binom{8}{1} (1)^7 (2x - 2x^2)^1 = \binom{8}{1}(2x - 2x^2)$ $= 8(2x - 2x^2) = 16x - 16x^2$

Third term ($k = 2$):

$\binom{8}{2} (1)^6 (2x - 2x^2)^2$ Expand $(2x - 2x^2)^2$: $(2x - 2x^2)^2 = 4x^2 - 8x^3 + 4x^4$ Only the terms up to $x^3$ matter, so: $\binom{8}{2} (4x^2 - 8x^3) = \frac{8 \cdot 7}{2} (4x^2 - 8x^3) = 28(4x^2 - 8x^3)$ $= 112x^2 - 224x^3$

Fourth term ($k = 3$):

$\binom{8}{3} (1)^5 (2x - 2x^2)^3$ Expand $(2x - 2x^2)^3$, but since it involves $x^4$ and higher, we ignore this term as it doesn't contribute to $x^3$.

Step 3: Combine terms

Add all contributions up to $x^3$: $1 + (16x - 16x^2) + (112x^2 - 224x^3)$ $= 1 + 16x + (112x^2 - 16x^2) - 224x^3$ $= 1 + 16x + 96x^2 - 224x^3$

Final Answer:

$1 + 16x + 96x^2 - 224x^3$

Would you like me to explain any part of this further or explore related concepts?

Here are 5 questions that build upon this explanation:

1. How does the binomial theorem help in polynomial expansion?
2. What happens if you need higher-order terms beyond $x^3$?
3. How would this change if the power of the binomial were larger (e.g., $n = 10$)?
4. What is the significance of combining like terms in the expansion process?
5. How would you approach the expansion if the expression had more complex coefficients?

Tip:

When working with expansions, keep track of terms systematically, especially when truncating at a specific degree!