MHB Jeffrrey's question at Yahoo Answers regarding binomial expansion

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The discussion revolves around finding the value of t in the binomial expansion of the expression (3x^2 - t/x)^6, specifically identifying the term independent of x, which is given as 2160. Using the binomial theorem, the relevant term is derived by setting the exponent of x to zero, leading to the conclusion that k must equal 4. The calculations show that the expression simplifies to 15 * 9 * t^4 = 2160, allowing for the determination that t^4 equals 14, resulting in t being 2. The final answer confirms that t is indeed 2, satisfying the conditions of the problem.
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Here is the question:

Can someone help me please?


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.
 
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Hello Jeffrey,

We are given the expression:

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

Using the binomial theorem, we may write:

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

We may rewrite this as:

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

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

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

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

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

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

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

$$t^4=14=2^4$$

For real $0<t$ we then find:

$$t=2$$
 
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