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Using the Rational Root Theorem, we find that $-1$ is a root. Since $x^4+3x^3+4x^2+3x+1$ is a reciprocal equation, we find that $\dfrac{1}{-1}=-1$ is another root, meaning that $-1$ is a double root. Then, dividing by $x^2+2x+1$ gives $x^2+x+1$ . Hence, $x^4+3x^3+4x^2+3x+1=(x+1)^2(x^2+x+1)$ Hence $a=b=1, c=\omega, d=\omega^2$ where $\omega$ and $\omega^2$ are the non-real cube roots of unity. Hence: $a^{1024}+b^{1024}+c^{1024}+d^{1024}+\frac{1}{a^{1024}}+\frac{1}{b^{1024}}+\frac{1}{c^{1024}}+\frac{1}{d^{1024}}\\=1+1+\omega^{1023}\times \omega+(\omega^2)^{1023}\times \omega^2+1+1+\frac{1}{\omega^{1023}\times \omega}+\frac{1}{(\omega^2)^{1023}\times \omega^2}\\=4+2\omega+2\omega^2=2(1+(1+\omega+\omega^2))=2\times 1=\boxed{2}$

$\begin{aligned} x^4+3x^3+4x^2+3x+1 & = 0 \quad \quad \small \color{#3D99F6}{\text{Divided throughout by }x^2} \\ x^2+3x+4+\frac{3}{x}+\frac{1}{x^2} & = 0 \\ \left(x+ \frac{1}{x} \right)^2 + 3 \left(x+ \frac{1}{x} \right) + 2 & = 0 \\ \left(x+ \frac{1}{x} + 1\right) \left(x+ \frac{1}{x} + 2 \right) \\ (x^2+2x+1) (x^2+x+1) & = 0 \\ (x+1)^2(x^2+x+1) & = 0 \\ \Rightarrow x & = \begin{cases} a = -1 \\ b = -1 \\ c = \omega \\ d = \omega^2 \quad \quad \small \color{#3D99F6}{\text{where }\omega \text{ and } \omega^2 \text{ are the complex third roots of unity.}} \end{cases} \end{aligned}$

Therefore,

$\displaystyle a^{1024} + b^{1024} + c^{1024} + d^{1024} + \frac{1}{a^{1024}} + \frac{1}{b^{1024}} + \frac{1}{c^{1024}} + \frac{1}{d^{1024}} \\ \displaystyle = (-1)^{1024} + (-1)^{1024} + \omega^{1024} + \omega^{2048} + \frac{1}{(-1)^{1024}} + \frac{1}{(-1)^{1024}} + \frac{1}{\omega^{1024}} + \frac{1}{\omega^{2048}} \quad \quad \small \color{#3D99F6}{\text{Note that } \omega^3 = 1} \\ \displaystyle = 1 + 1 + \omega + \omega^{2} + 1 + 1 + \frac{1}{\omega} + \frac{1}{\omega^{2}} \\ = 1 + 1 + \omega + \omega^{2} + 1 + 1 + \omega^2 + \omega \\ = 4 + 2(\omega + \omega^2) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{Since } \omega^2 + \omega + 1 = 0} \\ = 4 + 2(-1) = \boxed{2}$

divide by $x^2$ and substitute $y=x+\dfrac{1}{x}$ . solve for y and you get -1 or-2. note that $x^2+\dfrac{1}{x^2}=y^2-2$ and like this we can find all power of. we can see the recurence that no matter how greater the power of 2≠0, it will always be -1 or 2. so two roots give -1, other 2 give 2. the sum becomes 2*2+2(-1)=2

For those who do not know the complex cube roots of unity (like me), you can solve this with Newton's sums.

$x^4 + 3x^3 + 4x^2 + 3x + 1 = 0$

Notice that $1 - 3 + 4 - 3 + 1 = 0$

From this, we can know that $x = -1$ is a root of the equation.

Factorize $x + 1$ out and you get:

$(x + 1)(x^3 + 2x^2 + 2x + 1) = 0$

Now, notice again that $- 1 + 2 - 2 + 1 = 0$

From this, we know that the root $x = -1$ is repeated

Factorize it out again, and you have:

$(x + 1)^2 (x^2 + x + 1) = 0$

The discriminant for $(x^2 + x + 1) = 1^2 - 4(1)(1) = -3$

The last $2$ solutions for the equation are complex roots.

But first, substitute the 2 known roots $(c = d = -1)$ into the expression:

$a^{1024} + b^{1024} + c^{1024} + d^{1024} + \dfrac{1}{a^{1024}} + \dfrac{1}{b^{1024}} + \dfrac{1}{c^{1024}} + \dfrac{1}{d^{1024}}$

$= a^{1024} + b^{1024} + (-1)^{1024} + (-1)^{1024} + \dfrac{1}{a^{1024}} + \dfrac{1}{b^{1024}} + \dfrac{1}{(-1)^{1024}} + \dfrac{1}{(-1)^{1024}}$

$= a^{1024} + b^{1024} + \dfrac{1}{a^{1024}} + \dfrac{1}{b^{1024}} + 4$

$= a^{1024} + b^{1024} + \dfrac{a^{1024} + b^{1024}}{a^{1024}b^{1024}} + 4$

$= a^{1024} + b^{1024} + \dfrac{a^{1024} + b^{1024}}{(ab)^{1024}} + 4$

Now, let $a$ and $b$ be the roots of $(x^2 + x + 1)$ . From Vieta's formula, we get:

$a + b = -1$

$ab = 1$

$\implies(ab)^{1024} = 1$

Substitute this into the expression, and we have:

$a^{1024} + b^{1024} + \dfrac{a^{1024} + b^{1024}}{(ab)^{1024}} + 4$

$= a^{1024} + b^{1024} + a^{1024} + b^{1024} + 4$

$= 2(a^{1024} + b^{1024}) + 4$

This is where Newton's sums comes in.

$a + b = -1$

$a^2 + b^2 = (a + b)^2 - 2ab = (-1)^2 - 2(1) = -1$

$a^3 + b^3 = (a + b)(a^2 + b^2) - ab(a + b) = -1(-1) - 1(-1) = 2$

$a^4 + b^4 = (a + b)(a^3 + b^3) - ab(a^2 + b^2) = -1(2) - 1(-1) = -1$

$a^5 + b^5 = (a + b)(a^4 + b^4) - ab(a^3 + b^3) = -1(-1) - 1(2) = -1$

$a^6 + b^6 = (a + b)(a^5 + b^5) - ab(a^4 + b^4) = -1(-1) - 1(-1) = 2$

$a^7 + b^7 = (a + b)(a^6 + b^6) - ab(a^5 + b^5) = -1(2) - 1(-1) = -1$

$a^8 + b^8 = (a + b)(a^7 + b^7) - ab(a^6 + b^6) = -1(-1) - 1(2) = -1$

$a^9 + b^9 = (a + b)(a^8 + b^8) - ab(a^7 + b^7) = -1(-1) - 1(-1) = 2$

and the list goes on......

From this list, you can notice that the Newton's sums will have a repeating sequence of $-1, -1, 2$ . Don't believe? Try expanding further and you'll see the pattern holds true. Thus, we can conclude that

$a^n + b^n = \begin{cases} -1 & \quad n \equiv 1 \;\text{or}\;n \equiv 2 \quad(\text{mod} \quad3)\\ 2 & \quad n \equiv 0 \quad(\text{mod} \quad3) \end{cases}$

With this general expression, we can know the value of $(a^{1024} + b^{1024})$ by determining the remainder of $1024$ when divided by $3$ :

$n = 1024 = 341\dfrac{1}{3} \equiv 1 \quad(\text{mod}\quad3)$

Therefore,

$a^{1024} + b^{1024} = -1$

The final value of the expression is:

$2(a^{1024} + b^{1024}) + 4$

$=2(-1) + 4 = \boxed{2}$

Let the equation be $f(x)$ .

First use the Rational Root Theorem see that the possible roots are $\pm{1}$ . Plug them both in to see if they work. After calculations, it should be clear that $f(-1) = 0$ whilst $f(1) > 0$ .

Note that when $x^{4} \times f(\frac{1}{x}) = 0 \Rightarrow f(\frac{1}{x}) = 0$ . Therefore, if $f(x) = 0$ , $f(\frac{1}{x}) = 0$ . This works because the value of the coefficients are symmetrical.

Since $x = -1$ is a factor of $f(x)$ , $\frac{1}{x}$ = $\frac{1}{-1} = -1$ is also a factor of $f(x)$ .

After using long division to calculate $\frac{f(x)}{(x-1)^{2}}$ ,

$f(x) = (x-1)^{2}(x^{2}+x+1)$

We can use the Quadratic Formula to calculate the roots of $(x^{2}+x+1)$ .

$x = \frac{-1\pm\sqrt{1-4}}{2} = \frac{-1\pm\sqrt{3}i}{2}$

Now that the roots are found, we can plug them into the expression. $x = -1$ is fairly straight forward. $x = \frac{-1\pm\sqrt{3}i}{2}$ can be a bit of a pain if simply just plugged in and expanded.

Let $c = (\frac{-1+\sqrt{3}i}{2})^{1024}$ and $d = (\frac{-1-\sqrt{3}i}{2})^{1024}$ , (In retrospect, I just realized I overrode the value of $c$ and $d$ ... oh well.)

If we apply the Perfect square Identity to both of these equations, the value inside the bracket alternates between these two values (as according to the laws behind complex numbers): $\frac{-1+\sqrt{3}i}{2}$ and $\frac{-1-\sqrt{3}i}{2}$ .

For $c$ , let the exponent follow the formula of $2^{n}$ , when $n$ is an even number (2 because the exponent halves after you apply the perfect square identity), it will be $\frac{-1+\sqrt{3}i}{2}$ , and if it is odd, it will be $\frac{-1-\sqrt{3}i}{2}$ . As for $d$ , it is the same, but since it begins with a different parity, it will be the inverse of the case of $c$ .

Since the exponent value that we want (which is one) is simply just $2^{0}$ , which is the same parity as the starting value, the value inside the bracket stays the same. Thus,

$c = \frac{-1+\sqrt{3}i}{2}$ and $d = \frac{-1-\sqrt{3}i}{2}$ ,

We can move on to substituting these values, but by first simplifying the process.

$c + d + \frac{1}{c} + \frac{1}{d}$

$= \frac{c^{2}d+cd^{2}+(c+d)}{cd}$

$= \frac{cd(c+d)+(c+d)}{cd}$

$= \frac{(c+d)(cd+1)}{cd}$

$c + d = -1$ and $cd = 2$

Substituting them into $\frac{(c+d)(cd+1)}{cd}$ will give us $\boxed{-2}$ .

Therefore, we can plug these values into the expression, which will be $1 + 1 + 1 + 1 - 2 = \boxed{2}$ .

Roots can be found out by dividing with x^2 and substituting in eqn. Then its done and answer is 2

Take common (x+1) by splitting 3x^3 as 2+1 next coeff as 2+2 next as 2+1. Then from x^3+2x^2+x+1 take x^3+1 and2x^2+2x as groups. We can get x+1 as common again. Hence roots come as -1,-1,w,w^2. Thus its solved.

Roots are(-1,-1,w,w^2)