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Let there be this binomial: $$ (\sqrt{2} + \sqrt[3]{3})^{8}$$ How many rational terms are there in it's development?

I tought that the number of terms is given by n + 1 = 8 + 1 = 9, but that doesn't seem to be the answer.

Guilherme Almeida
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    Precisely, the question is "How many rational terms are in its development", while you simply know how many terms are in its development. – Luiz Cordeiro May 13 '16 at 23:43
  • Given that the terms $\sqrt{2}$ and $\sqrt[3]{3}$ are (rationally) incommensurate, to get a rational term in the expansion requires an even power of the first term and a power that's a multiple of three for the second term. How many of the nine terms in all are of that form? – hardmath May 13 '16 at 23:51
  • For example the term involving $a^2b^6$ in $(a+b)^8$ will give a rational value in this case, – coffeemath May 13 '16 at 23:51

2 Answers2

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HINT: The terms have the form

$$\binom8k\left(\sqrt2\right)^k\left(\sqrt[3]3\right)^{8-k}=\binom8k2^{k/2}3^{(8-k)/3}\;.$$

In order for this to be rational, the exponents $\frac{k}2$ and $\frac{8-k}3$ both have to be non-negative integers.

Brian M. Scott
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there are 9 terms, but not all are rational. Every term is of the form: $c_n 2^\frac{a}{2}3^\frac{b}{3}$

$c_n$ is a rational number and can be ignored.

$\frac{a}{2},\frac{b}{3}$ must be whole numbers.

And, $a + b = 8$

Doug M
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