Question

By how much is the sum of 8a 4 – 2a 2 b 2 + b 4 and –10a 4 + 3a 2 b 2 + 3b 4 greater than –2a 4 – 2a 2 b 2 + 4b 4 ?​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given : Sum of  8a⁴– 2a² b² + b⁴ and –10a⁴ + 3a² b² + 3b⁴

To Find : how much is the sum  greater than –2a⁴ – 2a² b² + 4b⁴

Solution:

Sum of  8a⁴– 2a² b² + b⁴ and –10a⁴ + 3a² b² + 3b⁴

= ( 8a⁴– 2a² b² + b⁴) + ( –10a⁴ + 3a² b² + 3b⁴ )

combine like terms

= 8a⁴ –10a⁴    – 2a² b² + 3a² b²  + b⁴ + 3b⁴

= –2a⁴   + a² b² + 4b⁴

–2a⁴   + a² b² + 4b⁴   is N greater than –2a⁴ – 2a² b² + 4b⁴

=> –2a⁴   + a² b² + 4b⁴  = N + ( –2a⁴ – 2a² b² + 4b⁴)

=> N = –2a⁴   + a² b² + 4b⁴ – ( –2a⁴ – 2a² b² + 4b⁴)

=> N =   –2a⁴   + a² b² + 4b⁴ + 2a⁴  + 2a² b² - 4b⁴

=> N =  3a² b²

Hence Sum of  8a⁴– 2a² b² + b⁴ and –10a⁴ + 3a² b² + 3b⁴   is   3a² b²

greater than  –2a⁴ – 2a² b² + 4b⁴

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