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

May 17, 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:

Binomial expansion of ${\left(a - b\right)}^{n}$ is given by

${C}_{0}^{n} {a}^{n} {b}^{0} + {C}_{1}^{n} {a}^{n - 1} {b}^{1} + {C}_{2}^{n} {a}^{n - 2} {b}^{2} + {C}_{3}^{n} {a}^{n - 3} {b}^{3} + \ldots \ldots + {C}_{r}^{n} {a}^{n - r} {b}^{r} + \ldots \ldots \ldots . + {C}_{n}^{n} {a}^{0} {b}^{n}$, where C_r^n=(n!)/((n-r)!r!).

Hence ${\left({x}^{2} + {y}^{2}\right)}^{6}$

= ${C}_{0}^{6} {\left({x}^{2}\right)}^{6} {\left({y}^{2}\right)}^{0} + {C}_{1}^{6} {\left({x}^{2}\right)}^{6 - 1} {\left({y}^{2}\right)}^{1} + {C}_{2}^{6} {\left({x}^{2}\right)}^{6 - 2} {\left({y}^{2}\right)}^{2} + {C}_{3}^{6} {\left({x}^{2}\right)}^{6 - 3} {\left({y}^{2}\right)}^{3} + {C}_{4}^{6} {\left({x}^{2}\right)}^{6 - 4} \left({y}^{2}\right) + {C}_{5}^{6} {\left({x}^{2}\right)}^{6 - 5} {\left({y}^{2}\right)}^{5} + {C}_{6}^{n} {\left({x}^{2}\right)}^{0} {\left({y}^{2}\right)}^{6}$

= ${x}^{12} + \frac{6}{1} \times {x}^{10} {y}^{2} + \frac{6 \cdot 5}{1 \cdot 2} {x}^{8} {y}^{4} + \frac{6 \cdot 5 \cdot 4}{1 \cdot 2 \cdot 3} {x}^{6} {y}^{6} + \frac{6 \cdot 5 \cdot 4 \cdot 3}{1 \cdot 2 \cdot 3 \cdot 4} {x}^{4} {y}^{8} + \frac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} {x}^{2} {y}^{10} + {y}^{12}$

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