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71
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71
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\left(\sqrt{62}\right)^{2}+6\sqrt{62}+9-6\sqrt{62}
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{62}+3\right)^{2}.
62+6\sqrt{62}+9-6\sqrt{62}
The square of \sqrt{62} is 62.
71+6\sqrt{62}-6\sqrt{62}
Add 62 and 9 to get 71.
71
Combine 6\sqrt{62} and -6\sqrt{62} to get 0.
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