Sure, let's solve the expression step by step.
We need to expand the expression [tex]\((3a^2 - 2)^3\)[/tex].
To do this, we can use the binomial theorem:
[tex]\[
(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k
\][/tex]
In this case, the expression is in the form [tex]\((x + y)^3\)[/tex] with [tex]\(x = 3a^2\)[/tex] and [tex]\(y = -2\)[/tex].
So, applying the binomial theorem:
[tex]\[
(3a^2 - 2)^3 = \sum_{k=0}^{3} \binom{3}{k} (3a^2)^{3-k} (-2)^k
\][/tex]
Let’s compute each term in the sum:
1. For [tex]\(k = 0\)[/tex]:
[tex]\[
\binom{3}{0} (3a^2)^{3-0} (-2)^0 = 1 \cdot (3a^2)^3 \cdot 1 = 1 \cdot 27a^6 = 27a^6
\][/tex]
2. For [tex]\(k = 1\)[/tex]:
[tex]\[
\binom{3}{1} (3a^2)^{3-1} (-2)^1 = 3 \cdot (3a^2)^2 \cdot (-2) = 3 \cdot 9a^4 \cdot (-2) = 3 \cdot 9 \cdot (-2) \cdot a^4 = -54a^4
\][/tex]
3. For [tex]\(k = 2\)[/tex]:
[tex]\[
\binom{3}{2} (3a^2)^{3-2} (-2)^2 = 3 \cdot (3a^2)^1 \cdot 4 = 3 \cdot 3a^2 \cdot 4 = 3 \cdot 3 \cdot 4 \cdot a^2 = 36a^2
\][/tex]
4. For [tex]\(k = 3\)[/tex]:
[tex]\[
\binom{3}{3} (3a^2)^{3-3} (-2)^3 = 1 \cdot (3a^2)^0 \cdot (-8) = 1 \cdot 1 \cdot (-8) = -8
\][/tex]
Now, summing all these terms together:
[tex]\[
27a^6 + (-54a^4) + 36a^2 + (-8)
\][/tex]
So, the expanded form of [tex]\((3a^2 - 2)^3\)[/tex] is:
[tex]\[
27a^6 - 54a^4 + 36a^2 - 8
\][/tex]
Therefore,
[tex]\[
(3a^2 - 2)^3 = 27a^6 - 54a^4 + 36a^2 - 8
\][/tex]
