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April 16th, 2015, 10:29 AM  #1 
Senior Member Joined: Jan 2015 From: USA Posts: 107 Thanks: 2  Expansion to the 9th power
Hi! The problem is in the image attached. I dont know how to solve these kind of problems involving expansions beyond the 4th power. Thanx! 
April 16th, 2015, 10:44 AM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs 
The binomial theorem tells us that: $\displaystyle \left(ax^2\frac{1}{ax}\right)^9=\sum_{k=0}^9\left({9 \choose k}\left(ax^2\right)^{9k}\left(\frac{1}{ax}\right)^k\right)=\sum_{k=0}^9\left({9 \choose k}(1)^ka^{92k}x^{3(6k)}\right)$ We see now that the term for which the exponent of $x$ is 9 has $k=3$: $\displaystyle {9 \choose 3}(1)^ka^{92\cdot3}x^{3(63)}$ Hence, we obtain: $\displaystyle 84a^3=\frac{21}{2}$ Solving for $a$, we find: $\displaystyle a^3=\frac{1}{8}$ $\displaystyle a=\frac{1}{2}$ 
April 16th, 2015, 10:48 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
What ever the degree, you should know that the binomial theorem says that $\displaystyle (a+ b)^n= \sum \frac{n!}{i!(n i)!} a^ib^{ni}$ Here, $\displaystyle a= ax^2$ and $\displaystyle b= \frac{1}{ax}$ So the first thing you need to figure out is "what value of i makes $\displaystyle (ax^2)^i(1/ax)^{9i}$ gives x the 9th power? 
April 16th, 2015, 12:57 PM  #4  
Senior Member Joined: Jan 2015 From: USA Posts: 107 Thanks: 2  Quote:
Thanx a lot! Also thanx for explaining the theory needed  

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