To simplify or expand the expression $(-3ab^2 - 7bc^2)^3$, let's follow the steps:

Step 1: Recall the Binomial Theorem

The Binomial Theorem states that for $(x + y)^n$, $(x + y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k} y^k,$ where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the binomial coefficient.

For $(-3ab^2 - 7bc^2)^3$, substitute $x = -3ab^2$, $y = -7bc^2$, and $n = 3$.

Step 2: Expand the Expression

$(-3ab^2 - 7bc^2)^3 = \sum_{k=0}^3 \binom{3}{k} (-3ab^2)^{3-k} (-7bc^2)^k.$

Compute each term:

Term 1: $k = 0$

$\binom{3}{0} (-3ab^2)^3 (-7bc^2)^0 = 1 \cdot (-3)^3 (ab^2)^3 \cdot 1 = -27a^3b^6.$

Term 2: $k = 1$

$\binom{3}{1} (-3ab^2)^2 (-7bc^2)^1 = 3 \cdot (-3)^2 (ab^2)^2 \cdot (-7)(bc^2) = 3 \cdot 9 \cdot (-7) \cdot a^2b^4c^2 = -189a^2b^4c^2.$

Term 3: $k = 2$

$\binom{3}{2} (-3ab^2)^1 (-7bc^2)^2 = 3 \cdot (-3)^1 (ab^2)^1 \cdot (-7)^2 (bc^2)^2 = 3 \cdot (-3) \cdot 49 \cdot ab^3c^4 = -441ab^3c^4.$

Term 4: $k = 3$

$\binom{3}{3} (-3ab^2)^0 (-7bc^2)^3 = 1 \cdot 1 \cdot (-7)^3 (bc^2)^3 = -343b^3c^6.$

Step 3: Combine the Terms

$(-3ab^2 - 7bc^2)^3 = -27a^3b^6 - 189a^2b^4c^2 - 441ab^3c^4 - 343b^3c^6.$

Final Answer:

$(-3ab^2 - 7bc^2)^3 = -27a^3b^6 - 189a^2b^4c^2 - 441ab^3c^4 - 343b^3c^6.$

Would you like further explanation on any step?

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5 Related Questions:

1. How do you calculate binomial coefficients efficiently?
2. Can this be visualized as a polynomial expansion in multiple variables?
3. How does the degree of the final polynomial relate to the original expression?
4. What happens if $a = 0$ or $b = 0$ in this polynomial?
5. Can this expansion be used in practical applications like physics or engineering?

Tip:

Always check your signs when raising negative terms to a power, as even and odd exponents alternate the sign!