b*ab = ab2
Suppose b*ab = ab + b2. Assume a and b are non-zero integers. Then
ab2 = ab + b2
b = 1 + b/a would have to be true for all b.
Counter-example:
b = 2; a = 3
b(ab) = 2(3)(2) = 12 = ab2 = (4)(3)
ab + b2 = (2)(3) + (2) = 10
but 10 does not = 12. Contradiction. So it cannot be the case that b = 1 + b/a is true for all b and, therefore, b*ab does not = ab + b2
Don't know
-2a^2
b2 + ab - 2 - 2b2 + 2ab = -b2 + ab - 2 which cannot be simplified further.
(a + x^2)(b + y^2)
ab=1a+1b a is equal to either 0 or two, and b is equal to a
That factors to (a + 1)(a + b) a = -1, -b b = -a
-2a^2
(a + x^2)(b + y^2)
b2 + ab - 2 - 2b2 + 2ab = -b2 + ab - 2 which cannot be simplified further.
ab=1a+1b a is equal to either 0 or two, and b is equal to a
(a+b)(a squared-ab+b squared)
x2y + axy + abx + a2b Factor by grouping. xy(x + a) + ab(x + a) (xy + ab)(x + a)
-132
No. A times B = AB
This expression can be factored. ab + 3a + b2 + 3b = a(b + 3) + b(b + 3) = (a + b)(b + 3)
That factors to (a + 1)(a + b) a = -1, -b b = -a
4
If, as is normal, ab represents a times b, etc then ab + ab + cc = 2ab + c2 which is generally not the same as abc.