Answer :
To solve the problem, we need to use the Binomial Theorem to expand the expression [tex]\((4y + 2x)^3\)[/tex]. The Binomial Theorem states that:
[tex]\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \][/tex]
For the expression [tex]\((4y + 2x)^3\)[/tex], we can identify:
- [tex]\(a = 4y\)[/tex]
- [tex]\(b = 2x\)[/tex]
- [tex]\(n = 3\)[/tex]
We will expand this expression by calculating each term where [tex]\(k\)[/tex] varies from 0 to [tex]\(n\)[/tex].
[tex]\[ (4y + 2x)^3 = \sum_{k=0}^{3} \binom{3}{k} (4y)^{3-k} (2x)^k \][/tex]
Let's compute each term individually:
### Term when [tex]\(k = 0\)[/tex]:
[tex]\[ \binom{3}{0} (4y)^{3-0} (2x)^0 = \binom{3}{0} (4y)^3 = 1 \cdot 64y^3 = 64y^3 \][/tex]
### Term when [tex]\(k = 1\)[/tex]:
[tex]\[ \binom{3}{1} (4y)^{3-1} (2x)^1 = \binom{3}{1} (4y)^2 (2x) = 3 \cdot 16y^2 \cdot 2x = 3 \cdot 32y^2x = 96y^2x \][/tex]
### Term when [tex]\(k = 2\)[/tex]:
[tex]\[ \binom{3}{2} (4y)^{3-2} (2x)^2 = \binom{3}{2} (4y) (2x)^2 = 3 \cdot 4y \cdot 4x^2 = 3 \cdot 16yx^2 = 48yx^2 \][/tex]
### Term when [tex]\(k = 3\)[/tex]:
[tex]\[ \binom{3}{3} (4y)^{3-3} (2x)^3 = \binom{3}{3} (4y)^0 (2x)^3 = 1 \cdot 8x^3 = 8x^3 \][/tex]
So the expansion of [tex]\((4y + 2x)^3\)[/tex] is:
[tex]\[ (4y + 2x)^3 = 64y^3 + 96y^2x + 48yx^2 + 8x^3 \][/tex]
To find the 4th term, we look at the sequence of terms obtained. The 4th term in this expansion is:
[tex]\[ 8x^3 \][/tex]
Hence, the 4th term is:
[tex]\[ 8x^3 \][/tex]
Thus, the correct answer among the given choices is:
[tex]\[ \boxed{8x^3} \][/tex]
[tex]\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \][/tex]
For the expression [tex]\((4y + 2x)^3\)[/tex], we can identify:
- [tex]\(a = 4y\)[/tex]
- [tex]\(b = 2x\)[/tex]
- [tex]\(n = 3\)[/tex]
We will expand this expression by calculating each term where [tex]\(k\)[/tex] varies from 0 to [tex]\(n\)[/tex].
[tex]\[ (4y + 2x)^3 = \sum_{k=0}^{3} \binom{3}{k} (4y)^{3-k} (2x)^k \][/tex]
Let's compute each term individually:
### Term when [tex]\(k = 0\)[/tex]:
[tex]\[ \binom{3}{0} (4y)^{3-0} (2x)^0 = \binom{3}{0} (4y)^3 = 1 \cdot 64y^3 = 64y^3 \][/tex]
### Term when [tex]\(k = 1\)[/tex]:
[tex]\[ \binom{3}{1} (4y)^{3-1} (2x)^1 = \binom{3}{1} (4y)^2 (2x) = 3 \cdot 16y^2 \cdot 2x = 3 \cdot 32y^2x = 96y^2x \][/tex]
### Term when [tex]\(k = 2\)[/tex]:
[tex]\[ \binom{3}{2} (4y)^{3-2} (2x)^2 = \binom{3}{2} (4y) (2x)^2 = 3 \cdot 4y \cdot 4x^2 = 3 \cdot 16yx^2 = 48yx^2 \][/tex]
### Term when [tex]\(k = 3\)[/tex]:
[tex]\[ \binom{3}{3} (4y)^{3-3} (2x)^3 = \binom{3}{3} (4y)^0 (2x)^3 = 1 \cdot 8x^3 = 8x^3 \][/tex]
So the expansion of [tex]\((4y + 2x)^3\)[/tex] is:
[tex]\[ (4y + 2x)^3 = 64y^3 + 96y^2x + 48yx^2 + 8x^3 \][/tex]
To find the 4th term, we look at the sequence of terms obtained. The 4th term in this expansion is:
[tex]\[ 8x^3 \][/tex]
Hence, the 4th term is:
[tex]\[ 8x^3 \][/tex]
Thus, the correct answer among the given choices is:
[tex]\[ \boxed{8x^3} \][/tex]
