Binomial Theorem...Please HELP!!?

Question

Find the term independent of x in the expansion of (2-x)^3 (1/3x - x)^6. Please include clear steps and explain as much as u can...wherever u can...Thanks!

What i mean the independent term of x is the term in the equation that doesnt have x. It is the constant. For example in the equation x^2 + 2x -15, -15 is the term independent of x.

Answer 1

You can always myltiply all the terms that means ex.
(2-x)^3=(2-x)*(2-x)*(2-x)
But that quite a task if the expression is powered to more than 2-3

Then you have the Binomial Formula for Positive integral n
. . . . . . k=n
(x+y)^n=∑ C(n,k)*(x^(n-k)*y^(k) ; (%%%%)
. . . . . . k=0

And C(n,k)=n!/[k!*(n-k)!]. .. . . . . . (######)

C(n,k) is called the binominial Coefficient
0!≡ 1 (Definition)
n!= n* (n-1)*......1

now your problem is reduced to find the binomial Cofficients for the expression (%%%%)

I will show you one abitray example:
C(6,4)=6!/[4!*(6-4)!]=
6 * 5 * 4 * 3 * 2 * 1 /[(4*3*2*1)*(2*1)]=
720/(24*2)=15

you can also use
Pascal triangel to find the Binomial Coefficients

n k →
↓
: . . . . . . . . . . . . . . .1
. . . . . . . . . . . . . . 1. 1
. . .. . .. . . . . . . . .1. 2 . 1
. . . . . . . . . . . . .1. 3 . 3 . 1
. . . . . . . . . . . .1. .4. 6. .4. 1
. . . . . . . . . . .1. .5.10.10. 5 .1
. . . . . . . . . .1.. 6.15.20 15 6. 1

It's easy to construct the triangel of Pascal,
you see the number under two other numbers is the sum of these numbers. It's is somtime helfuld to construct a P- triangel, if you don't have a calculator or a formula handbook, because you can also look the coefficient up in a table.

Take the example (1/3x - x)^6.we find by use of pascal T: the binomial coefficient to bee
1.. 6.15.20 15 6. 1 (the last line),
and we have from the formula (%%%%)

. . . . . . k=6
(1/3x - x)^6=∑ C(6,k)*(x^(6-k)*y^(k) ; (%%%%)
. . . . . . k=0
so C(6,0)=1, C(6,1)=6, C(6,2)=15, C(6,3)=20 etc

and your solution to this problem would be see (%%%%)
1 * (1/3x)^6 + 6*(1/3x)^5(-x)^1 + 15(1/3x)^4(-x)^2 + 20(1/x)^3(-x)^3 +15(1/3)^2(-x)^4 + 6^(1/3)^1(-x)^5 + 1*(-x)^6

Conclusion
1) you can alway do the hard task and calculate the oldfashion way
2) you can use Binomil formula and get the binomial coefficients 3 different way:
. . . . . . A) use Pascal Triangel
: . . . . . B) calculate by use of (#####)
: . . . . .C) or simply look the coefficient up in a formula
. . . . . . . . .book.

Answer 2

Well, I've been out of school for a while, but "independent of x" doesn't ring a bell for me. Is this a term where if you set an equation equal to zero, you describe which values of x are defined for that situation? If so, then the values of x in this case would be 2 and 0.

To determine this, all you need to do is find the values of x where the product would be zero. In the first part, a value of x = 2 will cause that to become zero (zero to any power is still zero). And in the second part, any other value than x = 0 would give you a non-zero result.

I hope this helps.

Answer 3

you can rewrite this expression noticing that (1/3x-x)^6=(1/9x^2-2/3+x^2)^3

(2-x)^3 (1/9x^2-2/3+x^2)^3

Ce only term independent of x is a constant term which does not contain x here (-4/3) ^3

Answer 4

Just mail me n I will tell u how to do it..cuz I cant explain it here..
See ya!