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 $a+b+c$ ?

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∻Solution by Steps∻ ‖ step 1 ⋮ To expand $(\sqrt{2}+\sqrt{3})^3$, we will use the binomial expansion formula $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. ‖ ‖ step 2 ⋮ Let $a=\sqrt{2}$ and $b=\sqrt{3}$. Compute $a^3$, $3a^2b$, $3ab^2$, and $b^3$. ‖ ‖ step 3 ⋮ Calculating $a^3 = (\sqrt{2})^3 = 2\sqrt{2}$. ‖ ‖ step 4 ⋮ Calculating $3a^2b = 3(\sqrt{2})^2\sqrt{3} = 3 \cdot 2\sqrt{3} = 6\sqrt{3}$. ‖ ‖ step 5 ⋮ Calculating $3ab^2 = 3\sqrt{2}(\sqrt{3})^2 = 3\sqrt{2} \cdot 3 = 9\sqrt{2}$. ‖ ‖ step 6 ⋮ Calculating $b^3 = (\sqrt{3})^3 = 3\sqrt{3}$. ‖ ‖ step 7 ⋮ Adding the results from steps 3 to 6, we get $(\sqrt{2}+\sqrt{3})^3 = 2\sqrt{2} + 6\sqrt{3} + 9\sqrt{2} + 3\sqrt{3}$. ‖ ‖ step 8 ⋮ Combine like terms to get $a\sqrt{2} + b\sqrt{3} = (2+9)\sqrt{2} + (6+3)\sqrt{3} = 11\sqrt{2} + 9\sqrt{3}$. ‖ ‖ step 9 ⋮ There is no $\sqrt{6}$ term, so $c=0$. ‖ ‖ step 10 ⋮ The values of $a$, $b$, and $c$ are $11$, $9$, and $0$ respectively. ‖ ‖ step 11 ⋮ To find $a+b+c$, we add $11+9+0$. ‖ ‖ step 12 ⋮ The sum is $11+9+0 = 20$. ‖ ∻Answer∻ ⚹ $a+b+c = 20$ ⚹ ∻Key Concept∻ ⚹Binomial Expansion⚹ ∻Explanation∻ ⚹The binomial expansion formula is used to expand expressions of the form $(a+b)^n$. In this case, we expanded $(\sqrt{2}+\sqrt{3})^3$ and combined like terms to find the coefficients $a$, $b$, and $c$.⚹◊What is the expansion of $(\sqrt{2}+\sqrt{3})^{3}$ in terms of $a \sqrt{2}+b \sqrt{3}+c \sqrt{6}$?⍭ Generate me a similar question◊

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