Radical Expression Sum 1
The number $ (\sqrt{2}+\sqrt{3})^3 $ can be written in the form $ a\sqrt{2} + b\sqrt{3} + c\sqrt{6} $, where $ a $, $ b $, and $ c $ are integers. What is $ a+b+c $?
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
First, we calculate $ (\sqrt{2} + \sqrt{3})^2 $: \begin{align*}
(\sqrt{2} + \sqrt{3})^2 &= (\sqrt{2} + \sqrt{3})(\sqrt{2}+\sqrt{3})\\
&=(\sqrt{2})(\sqrt{2}) + (\sqrt{2})(\sqrt{3}) + (\sqrt{3})(\sqrt{2}) + (\sqrt{3})(\sqrt{3})\\
&= 2 + \sqrt{6} + \sqrt{6} + 3\\
&=5+2\sqrt{6}.\end{align*} Multiplying this by $ \sqrt{2} +\sqrt{3} $ gives \begin{align*}
(\sqrt{2}+ \sqrt{3})^3 &=(\sqrt{2}+\sqrt{3})^2 (\sqrt{2} +\sqrt{3})\\
&=(5+2\sqrt{6})(\sqrt{2} +\sqrt{3})\\
&= 5\sqrt{2} + 5\sqrt{3} + (2\sqrt{6})(\sqrt{2}) + (2\sqrt{6})(\sqrt{3})\\
&=5\sqrt{2} + 5\sqrt{3} + 2\sqrt{12} + 2\sqrt{18}\\
&=5\sqrt{2} + 5\sqrt{3} + 2(2\sqrt{3}) + 2(3\sqrt{2})\\
&=11\sqrt{2} + 9\sqrt{3}.\end{align*} Therefore, we have $ a+b+c = \boxed{20} $. (Notice that $ c=0; $ tricky!)
We may also expand $ (\sqrt{3} + \sqrt{2})^3 $ using the Binomial Theorem, giving us $ {\sqrt{2}}^3 + 3{\sqrt{2}}^2\sqrt{3}+3\sqrt{2}{\sqrt{3}}^2+{\sqrt{3}}^3 $. Simplifying this yields $ 2\sqrt{2}+6\sqrt{3}+9\sqrt{2}+3\sqrt{3} = 11\sqrt{2}+9\sqrt{3} $, and once again $ a + b + c = \boxed{20} $.
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