as a2+b2=(a+ib)(a-ib), b2+2=[b - i sqrt(2)] [b + i sqrt(2)]
l a2 b2 is c2!!Its completely norma
A2 + B2 = C2 If C=8, then A2 + B2 = 64
(a3 + b3)/(a + b) = (a + b)*(a2 - ab + b2)/(a + b) = (a2 - ab + b2)
a2+b2+c2=x2+y2+z2 divide each side by 2 (a2+b2+c2)/2=(x2+y2+z2)/2 a+b+c=x+y+z
(a - b + 2)(a + b + 2)
(a+b+c) (a+b-c)
a2-b2 = (a-b)(a+b)
a2+2a2b+2ab2+b2
a2
as a2+b2=(a+ib)(a-ib), b2+2=[b - i sqrt(2)] [b + i sqrt(2)]
No. If you expand (a + b)2 you get a2 + 2ab + b2. This is not equal to a2 + b2
l a2 b2 is c2!!Its completely norma
(a + b)(a - b)
sqrt(a2 + b2) can't be simplified. Neither can (a2 + b2) .
a2-b2=(a-b)(a+b)
The reciprocal of a + bi is a - bi:1/(a + bi) since the conjugate is a - bi:= 1(a - bi)/[(a + bi)(a - bi)]= (a - bi)/[a2 - (b2)(i2)] since i2 equals to -1:= (a - bi)/(a2 + b2) since a2 + b2 = 1:= a - bi/1= a - bi