Question

expand (x + y)^6 =

Explanation:

Step1: Recall binomial theorem

The binomial theorem states that $(a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}$, where $\binom{n}{k}=\frac{n!}{k!(n - k)!}$. Here $a=x$, $b = y$ and $n = 6$.

Step2: Calculate binomial - coefficients and expand

When $k = 0$:
$\binom{6}{0}=\frac{6!}{0!(6 - 0)!}=1$, and $x^{6-0}y^{0}=x^{6}$, so the first term is $x^{6}$.
When $k = 1$:
$\binom{6}{1}=\frac{6!}{1!(6 - 1)!}=\frac{6!}{1!5!}=6$, and $x^{6 - 1}y^{1}=6x^{5}y$.
When $k = 2$:
$\binom{6}{2}=\frac{6!}{2!(6 - 2)!}=\frac{6\times5}{2\times1}=15$, and $x^{6-2}y^{2}=15x^{4}y^{2}$.
When $k = 3$:
$\binom{6}{3}=\frac{6!}{3!(6 - 3)!}=\frac{6\times5\times4}{3\times2\times1}=20$, and $x^{6 - 3}y^{3}=20x^{3}y^{3}$.
When $k = 4$:
$\binom{6}{4}=\frac{6!}{4!(6 - 4)!}=\frac{6\times5}{2\times1}=15$, and $x^{6-4}y^{4}=15x^{2}y^{4}$.
When $k = 5$:
$\binom{6}{5}=\frac{6!}{5!(6 - 5)!}=6$, and $x^{6 - 5}y^{5}=6xy^{5}$.
When $k = 6$:
$\binom{6}{6}=\frac{6!}{6!(6 - 6)!}=1$, and $x^{6-6}y^{6}=y^{6}$.

Answer:

$x^{6}+6x^{5}y + 15x^{4}y^{2}+20x^{3}y^{3}+15x^{2}y^{4}+6xy^{5}+y^{6}$