Is the hcf of a^3-8, a^4+16 and a^3+2a^2+4a just 1?

First we find the factors
a^3-8 is the differnce between 2 perfect cubes and is factored out to
(a-2)(a^2+2a+4)
the factors are (a-2) and (a^2+2a+4)

a^4+16
this is not factorable, unless you mistiped it and it was really a^4-16
the factor is 1

a^3+2a^2+4a
factor out the a
a(a^2+2a+4)
this is the most factored out form
the factors are (a) and (a^2+2a+4)

there are no common factors except 1 so 1 is the answer