Math, asked by anupam5624, 2 months ago

Long answer type question
1 By how much is a sum of a^4 -6a^2b^2+b^4 and -2a^4+5a^2 b^2+3b^4 greater than -a^4-a^2 b^2-4b^4 ?​

Answers

Answered by amitnrw
6

Given :   sum of a^4 -6a^2b^2+b^4 and -2a^4+5a^2 b^2+3b^4

To Find : How much it is greater than -a^4-a^2 b^2-4b^4 ?​

Solution:

sum of a⁴ -6a²b²+b⁴ and -2a⁴+5a² b²+3b⁴

=  a⁴ -6a²b²+b⁴ + ( -2a⁴+5a² b²+3b⁴)

= a⁴ -6a²b²+b⁴ -2a⁴+5a² b²+3b⁴

= -a⁴ -a²b² + 4b⁴

greater than -a⁴-a² b²-4b⁴

= -a⁴ -a²b² + 4b⁴ - (-a⁴-a² b²-4b⁴)

= -a⁴ -a²b² + 4b⁴ +  a⁴+a²b² + 4b⁴

= 8b⁴

sum of a⁴ -6a²b²+b⁴ and -2a⁴+5a² b²+3b⁴  is greater than -a⁴-a² b²-4b⁴  by 8b⁴

Learn More:

3p³q+2p²q+7 ; 2p²q+4pq-2p³q Add the given polynomial.

https://brainly.in/question/11760332

Answered by BrainlyAnswerer0687
3

Answer :

8b⁴

Solution in attachment

Attachments:
Similar questions