a) For commutative

Prove \(a\cdot b=b\cdot a\)

Now \(a\cdot b=a+b+2ab\)

and \(b\cdot a=b+a+2ba\)

\(b\cdot a=a+b=a+b+2ab\)

\(\displaystyle{\left\langle{a}+{b}={b}+{a}{\quad\text{and}\quad}{a}\cdot{b}={b}\cdot{a}\right\rangle}\)

Hence \(a\cdot b=b\cdot a\)

For associatvie

Prove\((a\cdot b)\cdot c=a\cdot(b\cdot c)\)

Now\((a\cdot b)\cdot c=(a+b+2ab)\cdot c\)

\(=a+b+2ab+c+2(a+b+2ab)c\)

\(=a+b+2ab+2ac+abc+4abc\)

and \(a\cdot (b\cdot c)=a\cdot (b+c+2bc)\)

\(=a+b+c+2bc+2a(b+c+2bc)\)

\(=a+b+c+2bc+2ab+2ac+4abc\)

\(=a+b+c+2ab+2ac+2bc+4abc\)

Hence \((a\cdot b)\cdot c=a\cdot(b\cdot c)\)