We need to use the following property of radicals to solve this question:
[tex]\sqrt{a\cdot b}=\sqrt{a}\sqrt{b}[/tex]The expression given is:
[tex]7\sqrt{3}+10\sqrt{108}[/tex]If we want to simplify, we need to rewrite the second term in terms of sqrt(3). We know that 108 is divisible by 3 because the sum of its digits is also divisible by 3.
Then:
[tex]\frac{108}{3}=36[/tex]We can rewrite:
[tex]7\sqrt{3}+10\sqrt{108}=7\sqrt{3}+10\sqrt{36\cdot3}[/tex]Using the property above:
[tex]7\sqrt{3}+10\sqrt{36}\sqrt{3}=7\sqrt{3}+10\cdot6\sqrt{3}[/tex]Now, we can simplify:
[tex]7\sqrt{3}+60\sqrt{3}=67\sqrt{3}[/tex]Thus, the answer is:
[tex]7\sqrt{3}+10\sqrt{108}=67\sqrt{3}[/tex]