To find the expanded form of the expression [tex]\((a - 3b)^3\)[/tex], we need to use the binomial theorem. The binomial theorem states that for any two numbers [tex]\(x\)[/tex] and [tex]\(y\)[/tex] and any non-negative integer [tex]\(n\)[/tex]:
[tex]\[
(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k
\][/tex]
In our case, we can rewrite [tex]\((a - 3b)^3\)[/tex] as:
[tex]\[
(a + (-3b))^3
\][/tex]
Applying the binomial theorem directly:
[tex]\[
(a - 3b)^3 = \sum_{k=0}^{3} \binom{3}{k} a^{3-k} (-3b)^k
\][/tex]
Expanding this sum, we get:
[tex]\[
\begin{align*}
(a - 3b)^3 &= \binom{3}{0} a^3 (-3b)^0 + \binom{3}{1} a^2 (-3b)^1 + \binom{3}{2} a^1 (-3b)^2 + \binom{3}{3} a^0 (-3b)^3 \\
&= 1 \cdot a^3 \cdot 1 + 3 \cdot a^2 \cdot (-3b) + 3 \cdot a \cdot (9b^2) + 1 \cdot 1 \cdot (-27b^3).
\end{align*}
\][/tex]
Next, we calculate each term individually:
1. [tex]\(a^3\)[/tex]:
[tex]\[
a^3 = a^3
\][/tex]
2. [tex]\(3 \cdot a^2 \cdot (-3b)\)[/tex]:
[tex]\[
3 \cdot a^2 \cdot (-3b) = -9a^2b
\][/tex]
3. [tex]\(3 \cdot a \cdot (9b^2)\)[/tex]:
[tex]\[
3 \cdot a \cdot (9b^2) = 27ab^2
\][/tex]
4. [tex]\(1 \cdot (-27b^3)\)[/tex]:
[tex]\[
-27b^3 = -27b^3
\][/tex]
Putting all these terms together, we get:
[tex]\[
(a - 3b)^3 = a^3 - 9a^2b + 27ab^2 - 27b^3
\][/tex]
To verify our result numerically with specific values for [tex]\(a\)[/tex] and [tex]\(b\)[/tex], let’s assume [tex]\(a = 1\)[/tex] and [tex]\(b = 1\)[/tex].
1. Calculate [tex]\(a^3\)[/tex]:
[tex]\[
1^3 = 1
\][/tex]
2. Calculate [tex]\(-9a^2b\)[/tex]:
[tex]\[
-9(1^2)(1) = -9
\][/tex]
3. Calculate [tex]\(27ab^2\)[/tex]:
[tex]\[
27(1)(1^2) = 27
\][/tex]
4. Calculate [tex]\(-27b^3\)[/tex]:
[tex]\[
-27(1^3) = -27
\][/tex]
Summing all these terms together:
[tex]\[
1 - 9 + 27 - 27 = -8
\][/tex]
Therefore, when [tex]\(a = 1\)[/tex] and [tex]\(b = 1\)[/tex], the expanded form [tex]\((a - 3b)^3\)[/tex] indeed simplifies to [tex]\(-8\)[/tex], as calculated.
