Big-Exponent Polynomial

Because Peter neglected powers of 3 and higher, this motivates me to factor the terms as much as possible in terms of product of cubic functions

The factorization of sum of cubes and difference of cubes is

$a^3 + b^3 = (a+b)(a^2-ab+b^2), \space a^3 - b^3 = (a-b)(a^2+ab+b^2)$ , so

$x^3 + 1 = (x+1)(x^2-x+1), \space x^3 - 1 = (x-1)(x^2+x+1)$

Group the terms as such

$\left ( (x+1)(x^2-x+1) \right )^{a+b} \cdot \left (- (x-1)(x^2+x+1) \right )^a \cdot (1+x)^c \cdot (1-x)^{b+c+d}$

$= (-1)^a \cdot (x^3 + 1)^{a+b} \cdot (x^3 - 1)^a \cdot (1-x^2)^c \cdot (1-x)^{b+d}$

Let $[[A], [B], [C], [D]]$ denote their respective powers of $x$ in $(x^3 + 1)^{a+b} , \space (x^3 - 1)^a , \space (1-x^2)^c , \space (1-x)^{b+d}$

Then coefficient of $x^1$ only occurs at $[[0],[0],[0],[1]]$ , then

$(-1)^{a} \cdot 1^{a+b} \cdot (-1)^a \cdot 1 \cdot (-1) (b+d) = -21 \Rightarrow b+d = 21$

And coefficient of $x^2$ only occurs at $[[0],[0],[2],[0]]$ and $[[0],[0],[0],[2]]$

$\left ( (-1)^a \cdot 1^{a+b} \cdot (-1)^a \cdot -c \cdot 1 \right ) + \left ( (-1)^a \cdot 1^{a+b} \cdot (-1)^a \cdot 1 \cdot (-1)^2 { b+d \choose 2 } \right ) = 0$

$\Rightarrow -c + (-1)^{2a} \cdot (-1)^2 {21 \choose 2 } = 0$

$\Rightarrow c = \boxed{210}$

We can first expand all the x terms. To do this, we can disregard all terms but the terms of degree \leq 1. We then have: (1-(a+b+c+d)x)(1+(a+b+c)x)(1-(a+b)x)(1+ax) => -(a+b+c+d)+(a+b+c)-(a+b)+(a) = -b-d = -21. Thus, b+d = 21. In a similar fashion, we can expand the x^2 terms to get the equation: (b^2+2bd+d^2-b-2c-d) = 0 (b+d)^2-(b+d) = 2c Using our result from the x term expansion: 441-21 = 2c c = 210 As desired.

We can rewrite the expression as: $(1-x^6)^a (1-x+x^3-x^4)^b(1-x^2)^c(1-x)^d$ Now we notice that expanding the product, the coefficient of $x^1$ will be: $-b-d$

this is due to the fact that we can have $x^1$ only by taking $-x$ from the factors: $(1-x+x^3-x^4)^b$ and $(1-x)^d$ and $1$ from all the other factors (and there are $b+d$ ways, but because we are taking $-x$ the coefficient will be the negative).

We know that this coefficient has to be $-21$ , therefore we get: $b+d=21$

the coefficient of $x^2$ will be: ${b+d \choose 2}-c$

this is due to the fact that we can have $-x^2$ only by taking $-x^2$ from the factor $(1-x^2)^c$ and $1$ from all the other factors (and there are $c$ ways, but because we are taking $-x^2$ the coefficient will be the negative).

Or by taking two times $-x$ from the factors $(1-x+x^3-x^4)^b$ and $(1-x)^d$ and $1$ from all the other factors (and there are ${b+d \choose 2}$ ways).

Knowing that the coefficient of $x^2$ has to be $0$ we get: ${b+d \choose 2}-c=0$ ${21 \choose 2}-c=0$ $210-c=0$ therefore $c$ is $\fbox{210}$

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

((1-x)^(a+b+c+d)) ((1+x)^(a+b+c)) (((1+x^3)/(1+x))^(a+b))*(((1-x^3)/(1-x))^(a))

=((1-x^2)^c) ((1-x)^(b+d)) ((1-x^6)^a)*((1+x^3)^b)

coefficient of x is -21, (b+d)C(1) (1)^(b+d-1) (-x)^1 = -21x --> b+d = 21

coefficient of x^2 is 0 (zero), (c)C(1) (1)^(c-1) (-x^2)^1 + (b+d)C(2) (1)^(b+d-2) (-x)^2 = 0 --> c=210

Rearranging $f$ slightly:

$\begin{aligned} f(x) = &(1 - x^2)^{a+b+c} \\ & (x^4 + x^2 - 1)^a \\ & (1-x+x^2)^b \\ &(1-x)^d \end{aligned}$

Expanding out the terms up to $x^2$ :

$\begin{aligned} f(x) = &(1 - (a+b+c)x^2 + ...) \\ &(1 + ax^2 + ...) \\ &(1 - bx + \frac{1}{2}b(b+1)x^2 - ...) \\ &(1 - dx + \frac{1}{2}d(d-1)x^2 - ...) \end{aligned}$

Expanding this and dropping $x^3$ and higher terms:

$\begin{aligned} f(x) &= 1 - (b+d)x \\ &+ (-a-b-c + a + \frac{1}{2}b(b+1) + \frac{1}{2}d(d-1) + bd)x^2 \end{aligned}$

The coefficient of $x$ gives us $d = 21-b$ .

The coefficient of $x^2$ :

$\begin{aligned} 0 &= -(a+b+c) + a + \frac{1}{2} (b(b+1) + (21-b)(20-b)) + b(21-b) \\ b+c &= \frac{1}{2} (2b^2 - 40b + 420) + 21b - b^2 \\ &= b + 210 \\ c &= \boxed{210} \end{aligned}$

We are going consider the coefficient of $x^2$ in the resulting polynomial.

We will now simplify the $f(x)$ .

$f(x)=(1-x)^{a+b+c+d}(1+x)^{a+b+c}(1-x+x^2)^{a+b}(1+x+x^2)^a$

$=[(1-x)^a(1+x)^a(1-x+x^2)^a(1+x+x^2)^a]$

$(1-x)^{b+c+d}[(1+x)^b(1-x+x^2)^b](1+x)^c$

$=(1-x^6)^a(1-x^3)^b[(1-x)^c(1+x)^c)](1-x)^{b+d}$

$=(1-x^6)^a(1-x^2)^c(1-x^3)^b(1-x)^{b+d}$

Firstly, consider the coefficient of $x$ .

Note that the only way to get a coefficient of $x$ is from the $(1-x)^{b+d}$ term, multiplied by the ones in the other terms.

This implies that $-21=-(b+d)$ , or $b+d=21$

Now, we will consider the coefficient of $x^2$ .

The first term and third terms, $(1-x^6)^a$ and $(1-x^3)^b$ must contribute a factor of 1 else the power of x is greater than 2.

Note that the coefficient of x is 0, since the result was $1-21x+0x^2+0x^3+0x^4$

So either the second term, $(1-x^2)^c$ , contributes an $x^2$ or fourth term, $(1-x)^{b+d}$ , contribute an $x^2$

Thus the coefficient of $x^2$ by multiplying out is:

$(-c)+\frac{(b+d)(b+d-1)}{2}=0$

Since we know $b+d=21$ , directly plugging it in results in $c=210$ .

Proof. Using the additivity of the exponents, we may factor out $\left(1-x\right)^{a}\left(1+x\right)^{a}\left(1-x+x^{2}\right)^{a}\left(1+x+x^{2}\right)^{a}=\left(1-x^{6}\right)^{a}$ from the product. We note that this term cannot affect any coefficients of $x$ with powers less than or equal to $4$ , so we discard it entirely without affecting the first five coefficients of our polynomial. This allows us to consider the product

$\left(1-x\right)^{b+d}\left(\left(1-x\right)\left(1+x\right)\right)^{c}\left(\left(1+x\right)\left(1-x+x^{2}\right)\right)^{b}$

$=\left(1-x\right)^{b+d}\left(1-x^{2}\right)^{c}\left(1-x^{3}\right)^{b}.$

The terms $\left(1-x^{2}\right)^{c}$ and $\left(1-x^{3}\right)^{b}$ play no role in the linear term, so it follows that $-21=-\binom{b+d}{1}x$ , and so $b+d=21$ . Looking at the quadratic term, our polynomial expands as $1-21x+\left(\binom{21}{2}-\binom{c}{1}\right)x^{2}+O\left(x^{3}\right)$ where by $O\left(x^{3}\right)$ , we mean all terms with degree $\geq3$ . Since the coefficient of $x^{2}$ must be zero, it follows that $c=\binom{21}{2}=210$ .

First notice that we have $[(1-x)(1+x)(1-x+x^2)(1+x+x^2)]^a*[(1-x)(1+x)(1-x+x^2)]^b*[(1-x)(1+x)]^c*[(1-x)]^d$ Simplifying we get $(1-x^6)^a*(-x^4+x^3-x+1)^b*(1-x^2)^c*(1-x)^d$ We can ignore any term which has power geq than 3, so we have $(1-x)^{b+d}(1-x^2)^c$ Now notice that in this, the coefficient of -x is b+d, so b+d = 21. Also, notice that there are two ways to get an $x^2$ , by timesing two -x's, or taking one x^2 from the second brackets. Since the ways of making x^2's from two -x's is (b+d)(b+d-1)/2 = 210, c has to also be $\boxed{210}$ to cancel it out to remove any x^2's. To explain for example why the coefficient of -x for example is b+d, imagine the expression written as all its brackets, i.e (1-x)(1-x)(1-x)...(1-x^2)(1-x^2)(1-x^2)... . Now to make a -x, there is only one thing we can do, take one from any of the b+d (1-x) brackets, and then times it by the 1's from the other brackets. Therefore we get b+d -x's. I used a similar idea to see what the other coefficients should be. I know this isn't that rigorous, but I can't think of a better way to put it.

Given \begin{equation} f(x) = (1−x) ^ {a+b+c+d} (1+x)^ {a+b+c} (1−x+x^2)^ {a+b} (1+x+x^2)^a \end{equation}

By the laws of exponents and rearranging the factors, \begin{equation} f(x) = (1−x) ^ a (1+x+x^2)^a (1+x)^ {a} (1−x+x^2)^ {a} (1+x)^ {b} (1−x+x^2)^ {b} (1+x)^ {c} (1−x) ^ {c} (1−x) ^ {b+d} \end{equation}

Applying special products, \begin{equation} f(x) = (1−x^6) ^ a (1+x^3)^ {b} (1-x^2)^ {c} (1−x) ^ {b+d} \end{equation}

Observe that all the terms, except the constant term, of the expansion of \begin{equation} (1−x^6) ^ a \end{equation} are all raised to a power of a multiple of 6. Thus, we ignore this factor.

Again, with the terms of \begin{equation} (1+x^3)^ {b} \end{equation} we get the factor

1. Sub polynomial multiplication - Multply (1+x)(1-x+x^2) =(1-x^3) and (1-x)(1+x+x^2)=(1-x^3)
2. You can now easily rule out expanding polynomial which is of higher degree
3. After applying the above step you can find that the coefficient of "x" is (b+d) which is 21 and the coefficient of "x^2" is (b+d)(b+d-1)/2 - c = 0
4. Hence concluded c = 210

We can rewrite the polynominal into $(1-x^{2})^{a+b+c}(1-x)^{d}(1+x^{2}+x^{4})^{a}(1-x+x^{2})^{b}$

(For easier calculation lol)

We can figure out that the coefficient of $x$ is $-d-b$ . (By the binominal theorem) Since the coefficient of $x$ is $-21$ , it gives us $b+d=21$ .

Now let's look at the coefficient of $x^{2}$ .

We can figure out that the coefficient of $x^{2}$ is $-c+_{d}C_{2}+_{b}C_{2}+bd$

It is 0, so $c$ equals to $_{d}C_{2}+_{b}C_{2}+bd$ .

And it's same with $\frac{d(d-1)}{2}+\frac{b(b-1)}{2}+bd=\frac{(b+d)^{2}}{2}-\frac{(b+d)}{2}$

Since we know that b+d=21, The answer is $\boxed{210}$

Sorry for poor English :)

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