# How do you use the binomial theorem to expand and simplify the expression (x^2+y^2)^6?

Jul 7, 2017

${\left({x}^{2} + {y}^{2}\right)}^{6} = {x}^{12} + 6 {x}^{10} {y}^{2} + 15 {x}^{8} {y}^{4} + 20 {x}^{6} {y}^{6} + 15 {x}^{4} {y}^{8} + 6 {x}^{2} {y}^{10} + {y}^{12}$

#### Explanation:

Recall the Binomial Theorem

${\left(a + b\right)}^{n} = {\sum}_{r = 0}^{n} \setminus \left(\begin{matrix}n \\ r\end{matrix}\right) \setminus {a}^{n - r} \setminus {b}^{r}$

Where:

 ( (n), (r) ) =""^nC_r = (n!)/((n-r)!r!

is the combinatorial, which are also the numbers from the $n t h$ row of Pascal's Triangle

Thus we have:

${\left({x}^{2} + {y}^{2}\right)}^{6} = {\sum}_{r = 0}^{6} \setminus \left(\begin{matrix}6 \\ r\end{matrix}\right) \setminus {\left({x}^{2}\right)}^{6 - r} \setminus {\left({y}^{2}\right)}^{r}$

$\text{ } = \left(\begin{matrix}6 \\ 0\end{matrix}\right) {\left({x}^{2}\right)}^{6} + \left(\begin{matrix}6 \\ 1\end{matrix}\right) {\left({x}^{2}\right)}^{5} \left({y}^{2}\right) + \left(\begin{matrix}6 \\ 2\end{matrix}\right) {\left({x}^{2}\right)}^{4} {\left({y}^{2}\right)}^{2} + \left(\begin{matrix}6 \\ 3\end{matrix}\right) {\left({x}^{2}\right)}^{3} {\left({y}^{2}\right)}^{3} + \left(\begin{matrix}6 \\ 4\end{matrix}\right) {\left({x}^{2}\right)}^{2} {\left({y}^{2}\right)}^{4} + \left(\begin{matrix}6 \\ 5\end{matrix}\right) \left({x}^{2}\right) {\left({y}^{2}\right)}^{5} + \left(\begin{matrix}6 \\ 6\end{matrix}\right) {\left({y}^{2}\right)}^{6}$

$\text{ } = \left(1\right) \left({x}^{12}\right) + \left(6\right) \left({x}^{10}\right) \left({y}^{2}\right) + \left(15\right) \left({x}^{8}\right) \left({y}^{4}\right) + \left(20\right) \left({x}^{6}\right) \left({y}^{6}\right) + \left(15\right) \left({x}^{4}\right) \left({y}^{8}\right) + \left(6\right) \left({x}^{2}\right) \left({y}^{10}\right) + \left(1\right) \left({y}^{12}\right)$

$\text{ } = {x}^{12} + 6 {x}^{10} {y}^{2} + 15 {x}^{8} {y}^{4} + 20 {x}^{6} {y}^{6} + 15 {x}^{4} {y}^{8} + 6 {x}^{2} {y}^{10} + {y}^{12}$