My Math Forum Expansion to the 9th power

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April 16th, 2015, 11:29 AM   #1
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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!
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 April 16th, 2015, 11:44 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,208 Thanks: 516 Math Focus: Calculus/ODEs The binomial theorem tells us that: $\displaystyle \left(ax^2-\frac{1}{ax}\right)^9=\sum_{k=0}^9\left({9 \choose k}\left(ax^2\right)^{9-k}\left(-\frac{1}{ax}\right)^k\right)=\sum_{k=0}^9\left({9 \choose k}(-1)^ka^{9-2k}x^{3(6-k)}\right)$ We see now that the term for which the exponent of $x$ is 9 has $k=3$: $\displaystyle {9 \choose 3}(-1)^ka^{9-2\cdot3}x^{3(6-3)}$ Hence, we obtain: $\displaystyle -84a^3=\frac{21}{2}$ Solving for $a$, we find: $\displaystyle a^3=-\frac{1}{8}$ $\displaystyle a=-\frac{1}{2}$ Thanks from matisolla
 April 16th, 2015, 11:48 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 $\displaystyle (a+ b)^n= \sum \frac{n!}{i!(n- i)!} a^ib^{n-i}$ 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 $\displaystyle (ax^2)^i(1/ax)^{9-i}$ gives x the 9th power? Thanks from matisolla
April 16th, 2015, 01:57 PM   #4
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Quote:
 Originally Posted by MarkFL The binomial theorem tells us that: $\displaystyle \left(ax^2-\frac{1}{ax}\right)^9=\sum_{k=0}^9\left({9 \choose k}\left(ax^2\right)^{9-k}\left(-\frac{1}{ax}\right)^k\right)=\sum_{k=0}^9\left({9 \choose k}(-1)^ka^{9-2k}x^{3(6-k)}\right)$ We see now that the term for which the exponent of $x$ is 9 has $k=3$: $\displaystyle {9 \choose 3}(-1)^ka^{9-2\cdot3}x^{3(6-3)}$ Hence, we obtain: $\displaystyle -84a^3=\frac{21}{2}$ Solving for $a$, we find: $\displaystyle a^3=-\frac{1}{8}$ $\displaystyle a=-\frac{1}{2}$

Thanx a lot! Also thanx for explaining the theory needed

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