Answer :
(a) is false. Consider the identity matrix, [tex]A=I[/tex]. Then
[tex](I+I^{-1})^3=(2I)^3=8I[/tex]
bu
[tex]I^3+I^{-1}=I+I=2I[/tex]
(b) is false. Since [tex]A[/tex] is invertible, we have
[tex]ABA^{-1}=B\implies (AB)(A^{-1}A)=BA\implies AB=BA[/tex]
but in general, matrix multiplication is non-commutative.
(c) is true. Let [tex]B=A^8[/tex]; then
[tex]B(A^{-1})^8=A^7(AA^{-1})(A^{-1})^7=A^7(A^{-1})^7[/tex]
[tex]B(A^{-1})^8=A^6(AA^{-1})(A^{-1})^6=A^6(A^{-1})^6[/tex]
and so on, down to
[tex]B(A^{-1})^8=AA^{-1}=I[/tex]
which is to say [tex]B^{-1}=(A^{-1})^8[/tex].
(d) is false. Consider [tex]A=-I[/tex], so that
[tex](-I)+I=0[/tex]
which is singular (i.e. determinant is zero), and thus not invertible.
(e) is true. It follows from the distributivity of matrix multiplication:
[tex](I+A)(I+A^{-1})=I^2+AI+IA^{-1}+AA^{-1}=I+A+A^{-1}+I=2I+A+A^{-1}[/tex]
(f) is false for the same reason as (b). Expanding the product gives
[tex](A+B)(A-B)=A^2+BA-AB-B^2[/tex]
but in general, [tex]AB\neq BA[/tex].