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How would one go about solving this?

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This is where i am stuck I am not even sure if I am on the right track, as you can that this is have to use nCr concept (Pascals triangle I believe) here

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2 Answers

up vote 2 down vote accepted

So you need the power of $x$ to be $8$. The power of $x$ is

$$24-3r-r=24-4r$$

Solving $24-4r=8$ Yields $r=4$.

Then, the coefficient is

$$\binom{8}{4}2.4^41.1^4$$

which can easely be calculated.

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so I was half way there? –  JackyBoi Oct 17 '12 at 3:39
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Each term in $\left(ax^3+\frac{b}x\right)^8$, before you collect terms, is a product of $8$ factors, each of which is either $ax^3$ or $\frac{b}x$. Suppose that in a given term you have $k$ factors of $ax^3$ and therefore $8-k$ factors of $\frac{b}x$; then the term is

$$\left(ax^3\right)^k\left(\frac{b}x\right)^{8-k}=a^kb^{8-k}x^{3k-(8-k)}=a^kb^{8-k}x^{4k-8}\;.$$

You want the coefficient of $x^8$, and $x^{4k-8}=x^8$ if and only if $4k-8=8$, $4k=16$, and $k=4$. In other words, the only terms that give you $x^8$ are those of the form

$$\left(ax^3\right)^4\left(\frac{b}x\right)^4\;.$$

From the binomial theorem you know that $(u+v)^8=\sum_{k=0}^8\binom8ku^kv^{8-k}$. In your problem $u=ax^3$, $v=\frac{b}x$, and you want the $k=4$ term; that’s

$$\binom84u^4v^4=\binom84\left(ax^3\right)^4\left(\frac{b}x\right)^4=\binom84a^4b^4x^4\;,$$ so the coefficient of $x^4$ is $$\binom84a^4b^4=70a^4b^4\;.$$

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just a silly question how did 8 and 4 in the bracket become 70? –  JackyBoi Oct 17 '12 at 3:38
    
@JackyBoi: $\binom84$ is the binomial coefficient $\frac{8!}{4!4!}=\frac{8\cdot7\cdot6\cdot5}{4\cdot3\cdot2\cdot1}=70$. –  Brian M. Scott Oct 17 '12 at 3:42
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