Prove that $(a,b)^2=(a^2,b^2,ab)$ I am trying to prove that $(a,b)^2=(a^2,b^2,ab)$ and was told that this follows from some very basic $\gcd$ laws. What am I not seeing?

Let
$$
c=\gcd(a,b),\quad a=a_1c,\quad b=b_1c,
$$
with $\gcd(a_1,b_1)=1$. Then
$$
\gcd(a^2,ab,b^2)=\gcd(a_1^2c^2,b_1^2c^2,a_1b_1c^2)=c^2\gcd(a_1^2,b_1^2,a_1b_1).
$$
Since $\gcd(a_1,b_1)=1$, it follows that $\gcd(a_1^2,b_1^2)=1$, and therefore
$$
\gcd(a^2,ab,b^2)=c^2=[\gcd(a,b)]^2.
$$
