# How to expand (x^2−2y)^6 using the binomial theorem?

Jan 16, 2018

See a solution process below

#### Explanation:

Here ${\textcolor{w h i t e}{x}}^{n}$C_r=[n!]/{r!(n-r)!}
Similarly you can do for any value of n and r except the negative ones , you will require to first make n(power to which it is raised) positive .

see below for solution
I am not using ${\textcolor{w h i t e}{x}}^{n}$C_r=[n!]/{r!(n-r)!} right now , i tried to use but in preview everything was messed , I think that would go a little difficult to understand . So , I'm using Pascal's Triangle .

Using this we get
${\left({x}^{2} - 2 y\right)}^{6} = {\left({x}^{2}\right)}^{6} + 6 \cdot {\left({x}^{2}\right)}^{5} \cdot \left(- 2 y\right) + 15 \cdot {\left({x}^{2}\right)}^{4} \cdot {\left(- 2 y\right)}^{2} + 20 \cdot {\left({x}^{2}\right)}^{3} \cdot {\left(- 2 y\right)}^{3} + 15 {\left({x}^{2}\right)}^{2} \cdot {\left(- 2 y\right)}^{4} + 6 \cdot \left({x}^{2}\right) \cdot {\left(- 2 y\right)}^{6}$
Now only calculation part is left . I hope you will be able to do it .

If you face any difficulty then let me know in comments , i'll add calculation part .

You can view a similar question(with answer) here