Answer :
To expand [tex]\((y + 2x^3)^3\)[/tex], we'll use the Binomial Theorem. The Binomial Theorem states:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
For our expression, [tex]\(a = y\)[/tex], [tex]\(b = 2x^3\)[/tex], and [tex]\(n = 3\)[/tex]. We will compute each term in the expansion where:
[tex]\[
\left(y + 2x^3\right)^3 = \sum_{k=0}^{3} \binom{3}{k} y^{3-k} (2x^3)^k
\][/tex]
Let's calculate each term:
1. For [tex]\(k = 0\)[/tex]:
[tex]\[
\binom{3}{0} y^{3-0} (2x^3)^0 = 1 \cdot y^3 \cdot 1 = y^3
\][/tex]
2. For [tex]\(k = 1\)[/tex]:
[tex]\[
\binom{3}{1} y^{3-1} (2x^3)^1 = 3 \cdot y^2 \cdot 2x^3 = 6y^2x^3
\][/tex]
3. For [tex]\(k = 2\)[/tex]:
[tex]\[
\binom{3}{2} y^{3-2} (2x^3)^2 = 3 \cdot y \cdot (2x^3)^2 = 3 \cdot y \cdot 4x^6 = 12yx^6
\][/tex]
4. For [tex]\(k = 3\)[/tex]:
[tex]\[
\binom{3}{3} y^{3-3} (2x^3)^3 = 1 \cdot 1 \cdot (8x^9) = 8x^9
\][/tex]
Now, combine all the terms to form the complete expansion:
[tex]\[
y^3 + 6y^2x^3 + 12yx^6 + 8x^9
\][/tex]
Thus, the correct answer is C) [tex]\(y^3 + 6y^2x^3 + 12yx^6 + 8x^9\)[/tex].
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
For our expression, [tex]\(a = y\)[/tex], [tex]\(b = 2x^3\)[/tex], and [tex]\(n = 3\)[/tex]. We will compute each term in the expansion where:
[tex]\[
\left(y + 2x^3\right)^3 = \sum_{k=0}^{3} \binom{3}{k} y^{3-k} (2x^3)^k
\][/tex]
Let's calculate each term:
1. For [tex]\(k = 0\)[/tex]:
[tex]\[
\binom{3}{0} y^{3-0} (2x^3)^0 = 1 \cdot y^3 \cdot 1 = y^3
\][/tex]
2. For [tex]\(k = 1\)[/tex]:
[tex]\[
\binom{3}{1} y^{3-1} (2x^3)^1 = 3 \cdot y^2 \cdot 2x^3 = 6y^2x^3
\][/tex]
3. For [tex]\(k = 2\)[/tex]:
[tex]\[
\binom{3}{2} y^{3-2} (2x^3)^2 = 3 \cdot y \cdot (2x^3)^2 = 3 \cdot y \cdot 4x^6 = 12yx^6
\][/tex]
4. For [tex]\(k = 3\)[/tex]:
[tex]\[
\binom{3}{3} y^{3-3} (2x^3)^3 = 1 \cdot 1 \cdot (8x^9) = 8x^9
\][/tex]
Now, combine all the terms to form the complete expansion:
[tex]\[
y^3 + 6y^2x^3 + 12yx^6 + 8x^9
\][/tex]
Thus, the correct answer is C) [tex]\(y^3 + 6y^2x^3 + 12yx^6 + 8x^9\)[/tex].
