# Jeffrrey's question at Yahoo! Answers regarding binomial expansion

#### MarkFL

Staff member
Here is the question:

Question: The term independent of x in the expansion of ((3x^2) - (t/x))^6 is 2160.

Given that t > 0, find the value of t.

Solution so far:

Formula - (n r) ((x)^n-r) ((y)^r)

So:

(6 r) ((3x^2)^6-r) (-t/x)^r

However, I don't know where to go from here. I am supposed to get:

x^12-3r

and then from that I am supposed to get r = 4

and then t = 2
I have posted a link there to this topic so the OP can view my work.

#### MarkFL

Staff member
Hello Jeffrey,

We are given the expression:

$$\displaystyle \left(3x^2-\frac{t}{x} \right)^6$$

Using the binomial theorem, we may write:

$$\displaystyle \left(3x^2+\left(-\frac{t}{x} \right) \right)^6=\sum_{k=0}^6\left[{6 \choose k}\left(3x^2 \right)^{6-k}\left(-\frac{t}{x} \right)^k \right]$$

We may rewrite this as:

$$\displaystyle \left(3x^2+\left(-\frac{t}{x} \right) \right)^6=\sum_{k=0}^6\left[{6 \choose k}3^{6-k}(-t)^kx^{3(4-k)} \right]$$

Now, in order for a term to be independent of $x$, we require the exponent on $x$ to be zero, which occurs for:

$$\displaystyle 3(4-k)=0\implies k=4$$

We are told this term is equal to $2160$, hence we have:

$$\displaystyle {6 \choose 4}3^{6-4}(-t)^4x^{3(4-4)}=2160$$

$$\displaystyle \frac{6!}{4!(6-4)!}\cdot3^2\cdot t^4=2160$$

$$\displaystyle 15\cdot9\cdot t^4=2160$$

$$\displaystyle t^4=14=2^4$$

For real $0<t$ we then find:

$$\displaystyle t=2$$