If a⊙b=(a+b)^2-2ab, which of the following is(are) true?

Note that (a+b)^2-2ab = a^2 + b^2
Therefore a⊙b = a^2 + b^2

Now look at the options:
Ⅰ. a⊙b=b⊙a is TRUE
because a^2 + b^2 is commutative

Ⅱ. (a⊙b)⊙c=a⊙(b⊙c) FALSE
LHS:
(a⊙b)⊙c = [a^2 + b^2]^2 + c^2 which results in a^4 +...
RHS:
a⊙(b⊙c) = a^2 + [[b^2 + c^2]^2] which results in a^2 +...
Therefore, without further calculation, we see that the orders of a do not match

Ⅲ. a⊙1=a^2+1 TRUE
Because a^2 + b^2 = a^2 + 1^2 = a^2 + 1

Therefore ANSWER [D] Ⅰand Ⅲ


A. Ⅰ only
B. Ⅱonly
C. Ⅲonly
D. Ⅰand Ⅲ
E. Ⅰ,Ⅱand Ⅲ

=>

a⊙b=(a+b)^2-2ab=a^2+2ab+b^2-2ab = a^2 + b^2

Statement I
b⊙a = b^2+a^2 = a^2 + b^2 = a⊙b
Therefore, statement I is true.

Statement II
(a⊙b)⊙c = (a^2+b^2) ⊙c = (a^2+b^2)^2 +c^2 = a^4 + 2a^2b^2 + b^4 + c^2
a⊙(b⊙c) = a⊙ (b^2+c^2) = a^2+ (b^2+c^2)^2 = a^2 + b^4 + 2b^2c^2 + c^4
We can easily find a counterexample.
If a = 1, b = 2 and c = 3, then
(a⊙b)⊙c = (1⊙2)⊙3 = (1^2+2^2) ⊙3 = 5⊙3 = 5^2 + 3^2 = 25 + 9 = 34 and
a⊙(b⊙c) = 1⊙(2⊙3) = 1⊙(2^2+3^2) = 1⊙13 = 1^2+13^2 = 1 + 169 = 170.
Thus, statement II is false.

Statement III
a⊙1 = a^2 + 1^2 = a^2 + 1
Therefore, statement III is true.

Therefore, the answer is D.

Answer: D