The prime factorization of 108 is
[tex]108 = 2^2\cdot 3^3[/tex]
This means that we have
[tex]\sqrt{-108} = \sqrt{(-1)\cdot2^2\cdot 3^3} = \sqrt{-1}\cdot\sqrt{2^2}\cdot\sqrt{3^3} = \sqrt{-1}\cdot\sqrt{2^2}\cdot\sqrt{3^2\cdot 3} = \sqrt{-1}\cdot\sqrt{2^2}\cdot\sqrt{3^2}\cdot\sqrt{3}[/tex]
We can simplify the squares and the square roots:
[tex]\sqrt{-1}\cdot\sqrt{2^2}\cdot\sqrt{3^2}\cdot\sqrt{3} = i\cdot 2\cdot 3 \cdot \sqrt{3} = 6i\sqrt{3}[/tex]
Similarly, we have
[tex]\sqrt{-3} = \sqrt{(-1)\cdot 3} = \sqrt{-1}\cdot\sqrt{3}=i\sqrt{3}[/tex]
So, the difference is
[tex]\sqrt{-108}-\sqrt{-3} = 6i\sqrt{3}-i\sqrt{3} = 5i\sqrt{3}[/tex]
