n(2a - b)(2a + b)(4a^2 - 2ab + b^2)(4a^2 + 2ab + b^2)
5
A^2-2ab+B^2 is actually (A+B)^2 AB squared is A^2B^2 or (AB)^2
(a-b)2 = a2 _ 2ab+b2
No. If you expand (a + b)2 you get a2 + 2ab + b2. This is not equal to a2 + b2
504
a^2 + b^2 + 2ab = (a + b)^2
(a + b) = a^2 + 2ab + b^2 (a - b)^2 = a^2 - 2ab + b^2 or you can work like this: [a + (-b)]^2 = a^2 + 2a(-b) + (-b)^2 (a - b)^2 = a^2 - 2ab + b^2
b/2=2ab/+b means112.5
n(2a - b)(2a + b)(4a^2 - 2ab + b^2)(4a^2 + 2ab + b^2)
n(2a - b)(2a + b)(4a^2 - 2ab + b^2)(4a^2 + 2ab + b^2)
It is a trinomial.The square of (a + b)^2 is a^2 + 2ab + b^2.
No, ab squared is not the same as 2ab. ab squared (ab^2) means multiplying ab by itself, resulting in a^2 * b^2. On the other hand, 2ab means multiplying 2 by a and then by b, resulting in 2ab. These two expressions are not equivalent because ab^2 involves squaring the variable b, while 2ab does not involve squaring any variables.
It is a trinomial.The square of (a + b)^2 is a^2 + 2ab + b^2.
Algebraically:((a+b) / 2) / ((a+b) / (2ab)) = 2ab(a + b) / 2(a + b) = (2a2b + 2ab2) / (2a + 2b) = abAs an example, let a = 2 and b = 3 then:i) (a + b) / 2 = (2 + 3) / 2 = 5/2ii) (a + b) / (2 * a * b) = (2 + 3) / (2 * 2 * 3) = 5 / 12Therefore i) is ab times larger than ii). (for these specific example numbers it will be 6 times larger)
5
A^2-2ab+B^2 is actually (A+B)^2 AB squared is A^2B^2 or (AB)^2