a3 + b3 + c3 + 2(a2)b + 2(b2)c + 2(a2)c + 2ab2 + 2(c2)b +abc
(a - b + 2)(a + b + 2)
a2 + b2 + c2 - ab - bc - ca = 0 => 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ca = 0 Rearranging, a2 - 2ab + b2 + b2 - 2bc + c2 + c2 - 2ca + a2 = 0 => (a2 - 2ab + b2) + (b2 - 2bc + c2) + (c2 - 2ca + a2) = 0 or (a - b)2 + (b - c)2 + (c - a)2 = 0 so a - b = 0, b - c = 0 and c - a = 0 (since each square is >=0) that is, a = b = c
If you mean what is b4 x b2, the answer is b6...
let binomial be (a + b)now (a+b)3 will be (a+b)(a+b)2 = (a+b)(a2 + 2ab+ b2) = a(a2+ 2ab+ b2) + b(a2 + 2ab+ b2) = a3+ 2a2b+ ab2 + a2b + 2ab2 + b3 = a3+ 2a2b+ ab2 + a2b + 2ab2 + b3 = a3 +3a2b + 3ab2 +b3 hope it helped... :D
0
(a^2 - b)(a^2 + b)(a^4 + b^2)
There are 3 main rules for expansion of algebraic expressions. They are as follows: 1) a2 _ b2 = (a-b) (a+b) 2) (a+b)2 = a2 + 2ab +b2 3) (a-b)2 = a2 - 2ab +b2
(a+b)(a-b) = a2 - b2 if we replace b by ib in this equation, then we get: (a+ib)(a-ib) = a2 - (ib)2 = a2 + b2 if z=a+ib then its conjuguate is a-ib and their product is a2 + b2
1 ------ a+b=4 2 ------ ab=2 ====> 3. b = 2/a Sub 3 into 1 ===> a + 2/a = 4 mutiply both sides by a ===> a2 +2 -4a = 0 use quadratic formula to find a ==> a= 2 +sqrt(2) or a' = 2- sqrt(2) use these two values of a to find a value for b using equation 3 ===> using a, b= 2-sqrt(2) and using a', b= 2+sqrt(2) hence a4 + b4 = (2+sqrt(2))4 + (2-sqrt(2))4 = 136 (for both values of a and b)
a3 + b3 + c3 + 2(a2)b + 2(b2)c + 2(a2)c + 2ab2 + 2(c2)b +abc
No. If you expand (a + b)2 you get a2 + 2ab + b2. This is not equal to a2 + b2
(a - b + 2)(a + b + 2)
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
Right Bank A, Left Bank B A1 B1 A4 B4 B2 A3 B3 A2
Right Bank A, Left Bank B A1 B1 A4 B4 B2 A3 B3 A2
Put simply, the equation for solving a cubic equation is x2 + 2ax +b = (x+a)2 + b-a2. This leads to x = -a +/- (a2 -b)1/2.