The question is a little unclear. If you mean: (A - B) x A2, then the answer is: A3 - (B x A2) (or, more simply, A3-BA2) If you mean: ((A - B) x A)2, then the answer is: (A2 - AB)2 which becomes A4 - 2A3B + A2B2
A^2-2ab+B^2 is actually (A+B)^2 AB squared is A^2B^2 or (AB)^2
Factor by grouping. x2y - xyb - abx + ab2 The first two can factor out an xy, so xy(x - b) The second two can factor out a -ab, so -ab(x - b) and we have xy(x - b) - ab(x - b) Since what is inside the parentheses is alike, we can be assured that we have factored correctly and now continue to group: ANS: (x - b)(xy - ab)
(a x b)^b =ab x b^2 =ab^3
a+b(a+B)=ab
You want: abc + ab Factor out the common terms which are "a" and "b" ab ( c + 1 )
The Wallenberg family through Investor AB
-2a^2
ab(ab) =2ab
A^2-2ab+B^2 is actually (A+B)^2 AB squared is A^2B^2 or (AB)^2
b2 + ab - 2 - 2b2 + 2ab = -b2 + ab - 2 which cannot be simplified further.
(a+b)(a squared-ab+b squared)
The length of its side squared.
e 4----------abb 4----------ebc 5----------cd 6----------aba 6----------ebE 4----------Ab
ab x ac = ab - ac
(a + x^2)(b + y^2)
(a-b)^2 doesn't have a numeric value since there are no numbers associated with it, but you can definitely expand it, as it represents a formula, instead of an actual numeric expression.(a-b)(a-b)= a^2-ab-ab+b^2= a^2-2ab+b^2 (which is actually the rule for expanding!)
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