a square plus 2ab plus b square
(a+b)cube = a cube + b cube + 3a square b + 3ab square
'a' minus 'b' whole cube is equal to 'a cube' minus 'b cube' minus (3 a square b ) plus (3 a b square) . .. .....thanks
Yes. The square of a whole number is always a whole number. For example, 3 squared is 9, so the square root of 9 is 3. What you never have, is the square root of a whole number being a fraction that is not a whole number. The square root of a whole number is either a whole number or an irrational number. For example, the square root of 2 is irrational, because there are no 2 whole numbers a and b such that a/b squared is 2. This is not terribly difficult to prove, but I have already said too much; I have answered your question.
a2+b2+c2-2ab+2bc-2ca
a square plus 2ab plus b square
(A+B)2 = (A+B).(A+B) =A2+AB+BA+B2 =A2+2AB+ B2 So the Answer is A + B the whole square is equal to A square plus 2AB plus B square. Avinash.
(a+b)cube = a cube + b cube + 3a square b + 3ab square
1.774225a
'a' minus 'b' whole cube is equal to 'a cube' minus 'b cube' minus (3 a square b ) plus (3 a b square) . .. .....thanks
(a-b)2 = a2 _ 2ab+b2
Yes. The square of a whole number is always a whole number. For example, 3 squared is 9, so the square root of 9 is 3. What you never have, is the square root of a whole number being a fraction that is not a whole number. The square root of a whole number is either a whole number or an irrational number. For example, the square root of 2 is irrational, because there are no 2 whole numbers a and b such that a/b squared is 2. This is not terribly difficult to prove, but I have already said too much; I have answered your question.
(a+b+c)²=a²+b²+c²+ 2ab+2bc+2ac
(a+b-c)2 = a2 + b2 +c2 +2ab - 2bc - 2ac
a2+b2+c2-2ab+2bc-2ca
(a+b)(a-b)
if B*B = a, then B is square root of a