Question 1208689
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Answer: <font color=red>81</font> + <font color=red>33</font>*sqrt(6)


Work Shown
{{{(a+b)^3 = a^3 + 3a^2b+3ab^2 + b^3}}} Using the <a href="https://en.wikipedia.org/wiki/Binomial_theorem">binomial theorem</a>


{{{(3+sqrt(6))^3 = 3^3 + 3*3^2*sqrt(6) + 3*3*(sqrt(6))^2 + (sqrt(6))^3}}} Plug in a = 3 and b = sqrt(6)


{{{(3+sqrt(6))^3 = 27 + 27*sqrt(6) + 54 + (sqrt(6))^2*sqrt(6)}}}


{{{(3+sqrt(6))^3 = 27 + 27*sqrt(6) + 54 + 6*sqrt(6)}}}


{{{(3+sqrt(6))^3 = (27+54) + (27*sqrt(6)+6*sqrt(6))}}}


{{{(3+sqrt(6))^3 = 81 + (27+6)*sqrt(6)}}}


{{{(3+sqrt(6))^3 = red(81) + red(33)*sqrt(6)}}}


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Here is another route which is a bit slower, but could be useful to know.
{{{(a+b)^2 = a^2+2ab+b^2}}} The FOIL rule can be used to confirm this or use the binomial theorem.


{{{(3+sqrt(6))^2 = 3^2 + 2*3*sqrt(6) + (sqrt(6))^2}}} Plug in a = 3 and b = sqrt(6)


{{{(3+sqrt(6))^2 = 9 + 6*sqrt(6) + 6}}}


{{{(3+sqrt(6))^2 = 15 + 6*sqrt(6)}}}



Then,
{{{(3+sqrt(6))^3 = (3+sqrt(6))(3+sqrt(6))^2}}}


{{{(3+sqrt(6))^3 = (3+sqrt(6))(15 + 6*sqrt(6))}}}


{{{(3+sqrt(6))^3 = x(15 + 6*sqrt(6))}}} Let x = 3+sqrt(6)


{{{(3+sqrt(6))^3 = x*15 + x*6*sqrt(6)}}}


{{{(3+sqrt(6))^3 = (3+sqrt(6))*15 + (3+sqrt(6))*6*sqrt(6)}}} Plug in x = 3+sqrt(6)


{{{(3+sqrt(6))^3 = 45 + 15*sqrt(6)+18*sqrt(6) + 6*sqrt(6)*sqrt(6)}}}


{{{(3+sqrt(6))^3 = 45 + 15*sqrt(6)+18*sqrt(6) + 36}}}


{{{(3+sqrt(6))^3 = red(81) + red(33)*sqrt(6)}}}
Therefore, <font color=red>81</font> and <font color=red>33</font> go in those blanks in the order mentioned.
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