solve: `(a^(2) + b^(2) + 2ab) - (a^(2) + b^(2) - 2ab)`

Remove the brackets and simplify : (a^(2)+b^(2) +2ab) -(a^(2)+b^(2)- 2ab)

Simplify (a - b) (a ^(2) + b ^(2) + ab) - (a +b) (a ^(2) +b ^(2) - ab)

( a - b) ^(2) + 2ab = ? A. a^(2) - b^(2) B. a^(2) + b^(2) C. a^(2) - 4ab + b^(2) D. a^(2) - 2ab + b^(2)

tan [ i log ((a - ib)/(a + ib )) ] is equal to : a) ab b) (2 ab)/( a ^(2) - b ^(2)) c) (a ^(2) - b ^(2))/( 2 ab) d) (2 ab)/( a ^(2) + b ^(2))

Factorise the using the identity a ^(2) - 2 ab + b ^(2) = (a -b) ^(2). 4a ^(2) - 4 ab + b ^(2)

If a and b are real and i=sqrt(-1) then sin[i ln((a+ib)/(a-ib))] is equal to 1) (2ab)/(a^(2)-b^(2)) 2) (-2ab)/(a^(2)-b^(2)) 3) (2ab)/(a^(2)+b^(2)) 4) (-2ab)/(a^(2)+b^(2))

Solve: (a - b )x + (a + b)y = a^(2) - 2ab - b^(2) and (a + b) (x + y) = a^(2) + b^(2)

tan (i log ((a-ib)/(a+ib))) = (i) ab (ii) (2ab)/(a^(2) -b^(2)) (iii) (a^( 2) -b^(2))/(ab) (iv) (2ab)/(a^(2)+b^(2))

If (x+1)/(x-1)=(a)/(b) and (1-y)/(1+y)=(b)/(a), then the value of (x-y)/(1+xy) is (2ab)/(a^(2)-b^(2)) (b) (a^(2)-b^(2))/(2ab) (c) (a^(2)+b^(2))/(2ab) (d) (a^(2)-b^(2)backslash)/(ab)