Question

Please help me

Answer 1

Find the term independent of x in (1 + x + 2x^3) ((3/2)x^2 - (1/3x))^9.

Answer 2

\[\huge (1+x+2x^{3})(\frac{ 3 }{ 2 }x^{2}-\frac{ 1 }{ 3x })^{9}\]

Answer 3

I'm not quite that advanced in math, but wouldn't you first distribute the exponent 9?

Answer 4

i don't think so

Answer 5

you are interested in the term with x^0

Answer 6

yes

Answer 7

i can handle a single bracket but not two of them

Answer 8

The same goes for me, but here is an idea:

First, find the term that is independent of \(x\) in \(1 \times \left(\dfrac{3}{2}x^2 - \dfrac{1}{3x}\right)^9\).

Then find the term that is independent of \(x\) in \(x \times \left(\dfrac{3}{2}x^2 - \dfrac{1}{3x}\right)^9\).

Then find the term that is independent of \(x\) in \(2x^2 \times \left(\dfrac{3}{2}x^2 - \dfrac{1}{3x}\right)^9\).

Finally, add all of them.

Answer 9

yes @ParthKohli

Answer 10

then note that there are no cases for the first

Answer 11

I'll use \(p(x)\) for the expression raised to the power 9.

For the first case, you want the \(x^0\) term for \(p(x)\). The second case, \(x^{-1}\). The third case, \(x^{-2}\).

Answer 12

Again, I've never studied the binomial theorem properly, so I can't bother actually solving the problem. >_>

Answer 13

@ParthKohli - you made a minor error in your third expression - the term outside the braces should be \(2x^3\) and not \(2x^2\)

Answer 14

the first one is manageable

Answer 15

Uh oh, the domino effect.

You need the \(x^{-3}\) term for the third case.

Answer 16

x^-1 in second
x^-3 in third

Answer 17

cool thanks @ParthKohli

Answer 18

Oh. So like this:
http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=7014d9722bc42de3f621eaad1da2615e&title=Simplifying%20Algebraic%20Expressions%20Calculator&theme=blue&i0=(1%2Bx%2B2x%5E3)(3%2F2x%5E2-1%2F3x)%5E9&podSelect=&includepodid=Input&includepodid=Result&podstate=Result__Step-by-step%20solution&showAssumptions=1&showWarnings=1

Answer 19

No problem. ^_^

But I'm still not convinced that there's no shortcut to this question.

Answer 20

I will get back if i find a better method