To expand [tex]\((a + b)^3\)[/tex], we can use the Binomial Theorem. The Binomial Theorem states that for any positive integer [tex]\( n \)[/tex]:
[tex]\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \][/tex]
In this case, [tex]\( n = 3 \)[/tex]. So, the formula becomes:
[tex]\[ (a + b)^3 = \binom{3}{0} a^3 b^0 + \binom{3}{1} a^2 b^1 + \binom{3}{2} a^1 b^2 + \binom{3}{3} a^0 b^3 \][/tex]
Now, let's calculate each term separately.
1. The first term where [tex]\( k = 0 \)[/tex]:
[tex]\[ \binom{3}{0} a^3 b^0 \][/tex]
[tex]\[ \binom{3}{0} = 1 \][/tex]
[tex]\[ a^3 b^0 = a^3 \cdot 1 = a^3 \][/tex]
So, the first term is [tex]\( a^3 \)[/tex].
2. The second term where [tex]\( k = 1 \)[/tex]:
[tex]\[ \binom{3}{1} a^2 b^1 \][/tex]
[tex]\[ \binom{3}{1} = 3 \][/tex]
[tex]\[ a^2 b^1 = a^2 b \][/tex]
So, the second term is [tex]\( 3a^2 b \)[/tex].
3. The third term where [tex]\( k = 2 \)[/tex]:
[tex]\[ \binom{3}{2} a^1 b^2 \][/tex]
[tex]\[ \binom{3}{2} = 3 \][/tex]
[tex]\[ a^1 b^2 = ab^2 \][/tex]
So, the third term is [tex]\( 3ab^2 \)[/tex].
4. The fourth term where [tex]\( k = 3 \)[/tex]:
[tex]\[ \binom{3}{3} a^0 b^3 \][/tex]
[tex]\[ \binom{3}{3} = 1 \][/tex]
[tex]\[ a^0 b^3 = 1 \cdot b^3 = b^3 \][/tex]
So, the fourth term is [tex]\( b^3 \)[/tex].
Putting these terms together, we get the expanded form:
[tex]\[ (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \][/tex]
Therefore, the expansion of [tex]\((a + b)^3\)[/tex] is:
[tex]\[ a^3 + 3a^2b + 3ab^2 + b^3 \][/tex]
