Answer:
yes
Step-by-step explanation:
hope it will help
(a^4 + 4b^4) ÷ (a^2 - 2ab + 2b^2)
= [(a^2 - 2ab + 2b^2) (a^2 + 2ab + 2b^2)] / (a^2 - 2ab + 2b^2)
= a^2+2ab+2b^2 =The answer
(a + b)^2 = a^2 + 2ab + b^2 => square of sums
(a - b)^2 = a^2 - 2ab + b^2 => square of deference
and of course one of most important ones:
a^2 - b^2 = (a - b)(a + b) => difference of squares
Best Answer: (a^4 + 4b^4) ÷ (a^2 - 2ab + 2b^2)
= [(a^2 - 2ab + 2b^2) (a^2 + 2ab + 2b^2)] / (a^2 - 2ab + 2b^2)
= a^2 + 2ab + 2b^2
a^4 + 4b^4 => i.e. 4a^2b^2 ,
a^4 + 4a^2b^2 + 4b^4 => a^2 + 2ab + b^2 = (a + b)^2, if : a = a^2 , b = 2b^2:
(a^2 + 2b^2)^2 = a^4 + 4a^2b^2 + 4b^4 => We can't add or subtract the value to the expression.
a^4 + 4a^2b^2 + 4b^4 - 4a^2b^2 =>
(a^2 + 2b^2)^2 - 4a^2b^2 =>
(a^2 + 2b^2 - 2ab)(a^2 + 2b^2 + 2ab) =>
(a^2 - 2ab + 2b^2) (a^2 + 2ab + 2b^2)
Greetings!
