Precalculus algebra: Greatest term in polynomial What is the term with the highest power in
$$(x^3-2)^{16}-(x^4 + 3)^{12}$$
The textbook answer is $-32x^{45}$, but not sure I understand the method used to handle such a problem. Most answered questions on the Internet assumes that there are no unknowns by specifying what x is, or contains too little information.
What I can gather, I need to find two things: the power of the largest term and then the coefficient of that term. As long as I know the power, I know how to get the coefficient and then I compare them between the two expansions.
To find the largest power, I need to compare one term and the following one to try to find a spot where the ratio switches from above 1 to below 1, and that is my maximum.
I start by writing down the general terms:
$$(x^3-2)^{16} = \sum_{k=0}^{16} = \binom{16}{k}(x^3)^k(-2)^{16-k}$$
$$(x^4+3)^{12} = \sum_{k=0}^{12} = \binom{12}{k}(x^4)^k(3)^{12-k}$$
Approach 1
Assume that $T_k$ is the largest term. Then
$$\frac{T_{k+1}}{T_k} = \frac{16-k+1}{k} \cdot \frac{-2}{x^3} = \frac{2k-34}{kx^3}$$
This means that as long as 
$$2k-34 > kx^3$$
or 
$$k > \frac{34}{2-x^3}$$
...the terms will continue to increase.
Not really sure how to go on from here, especially since k is expressed in terms of x. Am I even on the right track?
Approach 2 
Surely, the largest power of x in both parts of the original expression is $x^{48}$ where k equals 16 and 12, respectively. However, since there is a minus sign between the two parts of the original expression, these cancel. The second largest power in the left part occurs when k = 15, which is then $3 \cdot 15 = 45$ and for the other it is $4 \cdot 11 = 44$. To get the coefficient, we do the following:
$$16-k = 45 \Leftrightarrow k = 16-45 = -29$$
$$\binom{16}{-29}(x^3)^{-29}(-2)^{16--29} = \binom{16}{-29}(x^3)^{-29}(-2)^{45}$$
However, this does not seem to make much sense. I have never seen a negative number (e. g. -29) in a binomial expression like that before.
Where have I gone wrong and what are some other productive approaches to this question?
 A: Writing out binomial coefficients is correct approach here. As you have noticed, the leading terms (those with the highest power) in both $(x^3-2)^{16}$ and $(x^4+ 3)^{12}$ are equal to $x^{48}$ so they cancel out. The second term in the expansion of $(x^3-2)^{16}$ is $\binom {16} 1 (x^3)^{15}(-2)^1 = -32x^{45}$, and in the expansion of $(x^4+ 3)^{12}$ is $\binom{12} 1(x^4)^{11}3^1 = 36x^{44}$, so we can conclude that $$ (x^3-2)^{16} - (x^4+ 3)^{12} = -32x^{45} -36x^{44} + \text{something of lesser degree}$$ and you get your answer. What you did wrong in the second approach is that it is not $16 - k = 45$, but $3(16-k) = 45$, since you have $(x^3)^{16-k}$.
A: The answer you give would suggest that they are looking for the term with the highest power, rather than the term with the largest coefficient. This term comes from the first bracket, as the term with the highest power in the second bracket is $36x^{44}$, once you've cancelled the $x^{48}$ term.
A: Using binomial expansion, you see that the terms of greatest degree of $(x^3-2)^{16}$ and $(x^4+3)^{12}$ cancel out. 
The polynomial $(x^3-2)^{16}$ is a polynomial in $x^3$ and the polynomial $(x^4+3)^{12}$ is a polynomial in $x^4$, so the second largest term of $(x^3-2)^{16}$ has degree $45$ and the second largest term of $(x^4+3)^{12}$ has degree $44$. Therefore we can dismiss $(x^4+3)^{12}$ and focus our attention on $(x^3-2)^{16}$. 
The binomial expansion of $(x^3-2)^{16}$ starts with 
$$
x^{48} + \binom{16}{1} (-2)^1 x^{45} + \cdots
$$
therefore your answer is $-32 x^{45}$. 
Hope that helps,
