**Toolbox:**

- a binary operation * defined on Q in commutative if $ a*b=b*a$

Step1:

(i) $a*b=a-b\qquad a,b \in Q$

but $b*a =b-a \neq a-b \qquad a,b \in Q$

Hence * operation is not commutative

Step2:

(ii)$a*b =a^2+b^2 \qquad a,b \in Q$

$b *a=b^2+a^2=a^2+b^2$

[addition is commutaive in Q]

$a *b =b*a$

* operation is commutative

Step3:

(iii)$a*b =a+ab \qquad a,b \in Q$

But $b *a=b+ba$

$a*b \neq b*a$

* operation is not commutative

Step4:

(iv)$a*b=(a-b)^2\qquad a,b \in Q$

$b*a=(b-a)^2$

$=(-(a-b))^2$

$=(a-b)^2$

$a*b =b*a$

* operation is commutative

Solution:* operation defined by (ii) and (iv) are commutative