How do you multiply ${(2 b^{2} - 2 c^{2})}^{2}$ $(2b^2-2c^2)^2$ ?

Question

How do you multiply ${(2 b^{2} - 2 c^{2})}^{2}$ $(2b^2-2c^2)^2$ ?

1 Answer

Impact of this question

1707 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-19T05:37:12

$4 b^{4} - 8 b^{2} c + 4 c^{2}$ $4b^4-8b^2c+4c^2$

Explanation:

Given -

${(2 b^{2} - 2 c)}^{2}$ $(2b^2- 2c)^2$
$(2 b^{2} - 2 c) (2 b^{2} - 2 c)$ $(2b^2- 2c)(2b^2- 2c)$

$2 b^{2} (2 b^{2}) - 2 c (2 b^{2}) + 2 b^{2} (- 2 c) - 2 c (- 2 c)$ $2b^2(2b^2)-2c(2b^2)+2b^2(-2c)-2c(-2c)$

$4 b^{4} - 4 b^{2} c - 4 b^{2} c + 4 c^{2}$ $4b^4-4b^2c-4b^2c+4c^2$

$4 b^{4} - 8 b^{2} c + 4 c^{2}$ $4b^4-8b^2c+4c^2$

Science

Math

Humanities

... and beyond

How do you multiply ${(2 b^{2} - 2 c^{2})}^{2}$ $(2b^2-2c^2)^2$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you multiply (2b2−2c2)2(2b^2-2c^2)^2?

1 Answer

Explanation:

Related questions

Impact of this question

How do you multiply ${(2 b^{2} - 2 c^{2})}^{2}$ $(2b^2-2c^2)^2$ ?