Answer:
-30
Step-by-step explanation:
You can work this two ways. One way is to factor √3 from both binomials. We'll show that second. The other way is to multiply it out in the usual way.
Multiply it out
[tex](\sqrt{3}+3\sqrt{27})(2\sqrt{3}-\sqrt{27})\\\\=(\sqrt{3})(2\sqrt{3})+(\sqrt{3})(-\sqrt{27})+(3\sqrt{27})(2\sqrt{3})+(3\sqrt{27})(-\sqrt{27})\\\\=2\sqrt{9}-\sqrt{81}+6\sqrt{81}-3\sqrt{27^2}\\\\=2\cdot 3-9+6\cdot 9-3\cdot 27\\\\=6-9+54-81=-30[/tex]
Simplify it first
[tex](\sqrt{3}+3\sqrt{27})(2\sqrt{3}-\sqrt{27})\\\\=(\sqrt{3}(1+3\sqrt{9}))(\sqrt{3}(2-\sqrt{9}))\\\\=3(1+3\cdot 3)(2-3)=3(10)(-1)=-30[/tex]
To me, the latter method seems easier.
