Given A2=A and I is the identity matrix of the same order as A.
Since A2=A, any occurrence of A2 can be replaced with A.
Expand (I+2A)2:
(I+2A)2=(I+2A)(I+2A)
=I⋅I+I⋅2A+2A⋅I+2A⋅2A
=I+2A+2A+4A2
=I+2A+2A+4A
=I+8A
Expand (I+2A)3:
(I+2A)3=(I+2A)2×(I+2A)
=(I+8A)(I+2A)
=I⋅I+I⋅2A+8A⋅I+8A⋅2A
=I+2A+8A+16A2
=I+2A+8A+16A
=I+26A
Calculate (I+2A)3−6A:
(I+2A)3−6A=(I+26A)−6A
=I+26A−6A
=I+20A
Therefore, (I+2A)3−6A=I+20A