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Hey I am supposed to solve the following problem:

Specify a development member that does not contain an irrational number: $$\left (\sqrt{5} - \sqrt[3]{2} +2 \right )^{6}.$$

So I used multinomial theorem: $$\sum_{i=0}^{6}\binom{6}{n_{1},n_{2},n_{3}}\left ( \sqrt{5} \right )^{n_{1}}\left ( - \sqrt[3]{2} \right )^{n_{2}}\left ( 2 \right )^{n_{3}}$$ and then I know that :$ 2k+3l +n_{3}=6$ with $n_1=2k$ and $n_2=3l$.

Is that a correct answer?

$$\binom{6}{2,3,1}+\binom{6}{2,0,4}+\binom{6}{4,0,2}+\binom{6}{6,0,0}+\binom{6}{0,3,3}+\binom{6}{0,0,6}$$

Robert Z
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    A term will be rational if the $n_1=2k$ and $n_2=3l$, $k,l\in\mathbb{N}_0$. So you must find the non-negative integer solutions to $2k+3l+n_3=6$ – gustaffIR Jan 03 '20 at 15:58
  • @gustaffIR I edited my post, can you check it? –  Jan 03 '20 at 16:11

2 Answers2

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Hint: To get rid of the roots, you need $n_1$ be to even and $n_2$ to be a multiple of $3$.

lhf
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No, your answer is not correct. Since $n_1\in\{0,2,4,6\}$ and $n_2\in\{0,3,6\}$ with $n_1+n_2\leq 6$, we find that the required number is $$N:=\sum_{(n_1,n_2)\in S}\binom{6}{n_{1},n_{2},6-n_1-n_2}\left ( \sqrt{5} \right )^{n_{1}}\left ( - \sqrt[3]{2} \right )^{n_{2}}\left ( 2 \right )^{6-n_1-n_2}$$ where $S:=\{(0,0), (0,3), (0,6), (2,0), (2,3),(4,0), (6,0)\}$. What is $N$?

Robert Z
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