Complicated System of Equations

There's probably a simpler way, but here's a brute-force approach.

Solve the first equation for y, then plug it into the second equation and simply, you will get an expression of the form $f(x) = \pm (8-4x^3) \sqrt{5-4x^3}$ where f(x) is cubic in x. Square both sides, which preserves the number of solutions due to the plus-or-minus, and then simplify to get a degree 9 polynomial. The absolute value of the product of the roots is then the absolute value of the constant term divided by the leading term.

The only thing left to check is that there are no double roots, which can be done by checking that the GCD of the polynomial and its derivative is 1.

From the first equation, $y^2+y+(x^3-1)=0$ and solving the quadratic equation in $y$ , we get $y=\frac{-1 \pm \sqrt{1^2-4(x^3-1)}}{2} = \frac{-1 \pm \sqrt{5-4x^3}}{2}$ .

We then substitute this value of $y$ into the second equation. We get $x^2+(\frac{-1 \pm \sqrt{5-4x^3}}{2})^3 = 2 \Rightarrow 8x^2 + (-1 \pm \sqrt{5-4x^3})^3 - 16 = 0$ $\Rightarrow 8x^2 + ((-1) \pm 3\sqrt{5-4x^3} - 3(5-4x^3) \pm (5-4x^3)\sqrt{5-4x^3}) - 16 = 0$ and bringing the square root terms to one side we get $12x^3+8x^2-32 = \pm(8-4x^2)\sqrt{5-4x^3}$ $\Rightarrow 3x^3+2x^2-8=\pm(2-x^3)\sqrt{5-4x^3}$ $\Rightarrow (3x^3+2x^2-8)^2=(\pm(2-x^3)\sqrt{5-4x^3})^2$ $\Rightarrow (3x^3+2x^2-8)^2=(2-x^3)^2(5-4x^3)$

Note that if we expand and shift the terms to one side, we get a degree 9 polynomial with a leading coefficient of 4 and a constant term of 44. As the complex solutions are required, no extra or wrong solutions are generated from the above algebraic manipulations and thus the product of the possible $x$ values is the product of the roots of the polynomial so using Vieta's Formula, the required answer is $|-\frac{44}{4}|=11$

$$ x^3+y^2+y=1\ \ (1) \wedge x^2+y^3 =2\ \ (2) \Rightarrow y^3=2-x^2 $$ Consider (1): $$ (1)\Leftrightarrow x^3+y^2+y+1=2\Leftrightarrow x^3+\dfrac{y^3-1}{y-1}=2 $$ $$ \Leftrightarrow x^3+\dfrac{1-x^2}{y-1}=2 \Leftrightarrow y-1=\dfrac{1-x^2}{2-x^3} \Leftrightarrow y=\dfrac{3-x^2-x^3}{2-x^3} $$ Hence $$2-x^2=y^3=\left( \dfrac{3-x^2-x^3}{2-x^3} \right)^3 $$ $$\Leftrightarrow (x^3+x^2-3)^3+(2-x^2)(2-x^3)^3=0\ \ (*)$$ Using Vieta's theorem, suppose $$x 1,x 2,...,x n$$ are the roots of the system of equation given, we have $$|x 1x 2...x n|=-((-3)^3+2^4)=11.$$

x^8-x^7+x^6-4 x^5+7 x^4-6 x^3+3 x^2-11*x+11=0

We know that for any complex number $z, z.\overline{z}=|z|^2$ . Again, by the complex conjugate root theorem if $z$ is a root of the system then $\overline{z}$ is also a root. Also, $|abc...|=|a|.|b|.|c|...$ . So, basically the problem boils down to: "Eliminate $y$ , express the system solely as polynomial, $f$ on $x$ and the product of the roots is our answer." Using familiar factorization, $y^3-1=(y-1)(y^2+y+1)$ , we get, $f(x)=x^{11}-x^9-3x^8+3x^7+4x^6-6x^5-9x^4+3x^3+19x^2-11=0$ But there's a catch; multiple roots ! If $\exists$ multiple roots then $f(\alpha)=f'(\alpha)=0$ . Using this, we can get we have no multiple roots. So answer is $\boxed{11}$ since the last coefficient gives the product of the roots.

Slightly manipulate the first equation, we have $y^2+y+1=2-x^3$ . Now we can multiply both sides by $(y-1)$ to get $y^3-1=(2-x^3)(y-1)$ , which is much easier to work with than the original first equation. The equation can then be further manipulated into $y^3-1=(2-x^3)y-2+x^3$

From the second equation, we have $y^3=2-x^2$ , or $y=\sqrt[3]{2-x^2}$ . Substitute this back into the equation from above, we have $1-x^2=(2-x^3)\sqrt[3]{2-x^2}-2+x^3$ , or $3-x^2-x^3=(2-x^3)\sqrt[3]{2-x^2}$ . Cube both sides of the equation, and moving everything to the left to have $(3-x^2-x^3)^3-(2-x^3)^3(2-x^2)=0$

We can see the left side is a 11th degree polynomial. As we are looking for the product of roots, we are interested in the leading coefficient, which is $-1$ , and the constant term, which is $27-16=11$ . Thus the product of the roots is 11.

First we show that $y\neq{0}$ . If $y=0$ clearly the equations are not satisfied.

Now the idea is to extract from the above equations a polynomial of x, the roots of which comprise the set of x solutions to the system. To that purpose, we multiply the first equation by y.

$x^{3}+y^{2}+y=1\Rightarrow{x^{3}y+y^{3}+y^{2}=y}$

$y^{3}=2-x^{2}$

$y^{2}=1-y-x^{3}$ .

Substituting the last 2 expressions into the first, we obtain a relation that is linear in terms of y, and therefore easily solvable.

$y=\frac{x^{3}+x^{2}-3}{x^3-2}$ .

Upon back substitution, the desired polynomial for x, in implicit form is:

$x^{2}+\left(\frac{x^{3}+x^{2}-3}{x^3-2}\right)^{3}=2$

After verifying that $x^{3}\neq{2}$ , we have that:

$x^{2}(x^{3}-2)^{3}+(x^{3}+x^{2}-3)^{3}=2(x^{3}-2)^{3}$

We know that the absolute product of all roots of every polynomial is the absolute value of its constant term. In this case, we can easily calculate that the constant term is -11. So the answer is 11.

Note: if $x^{3}=2$ then

$y^{2}+y+1=0\Rightarrow{y=\frac{-1\pm{i\sqrt{3}}}{2}}$ , which does not satisfy both equations.

From the second equation, it can be shown that $y=\left(2-x^2\right)^{1/3}$ . Plugging this into the first equation yields $\left(2-x^2\right)^{2/3}+\left(2-x^2\right)^{1/3} =$ $1-x^3$ (let this be equation 1). Factoring on the left-hand side yields $\left(2-x^2\right)^{1/3}\left(\left(2-x^2\right)^{1/3}+1\right)=$ $1-x^3$ . Both sides are then cubed: $\left(2-x^2\right)\left(-x^2+3\left(2-x^2\right)^{2/3}+3\left(2-x^2\right)^{1/3}+3\right) =$ $\left(1-x^3\right)^3$ . But, from equation 1, substitution yields $-x^2+3\left(2-x^2\right)^{2/3}+3\left(2-x^2\right)^{1/3}+3 =$ $-x^2+3\left(1-x^3\right)+3 =$ $6 - x^2 - 3x^3$ . So all that is left is finding the product of the roots of $\left(2-x^2\right)\left(6-x^2-3x^3\right) =$ $\left(1-x^3\right)^3$ . Expansion yields the polynomial $0=x^9-3x^6+3x^5+x^4-3x^3-8x^2+11$ . From Vieta's formula, the absolute value of the product of the roots is $11/1 = 11$ .

x^3+y^2+y=1, find y in terms of x, y=(-1+sqrt(5+4x^3))/2 or y=(-1-sqrt(5+4x^3)/2 but 2nd one is the soln because at x=1 ,y=0,so real soln will be there. now substiute the y in x^2+y^3=2, now construct a polynomial in x, and hence product of roots =11

x^3+y^3=2 so y=(2-x^3)^(1/3) by putting this value in equation 1 we got x^3+(2-x^3)^(2/3)+(2-x^3)^(1/3)=1 (2-x^3)^(2/3)+(2-x^3)^(1/3)=1-x^3 cube both side & after simplifyng expression we get x^9+x^6-11x^3+11=0 so product of all x is 11

x^3+y^2+y=1 is a quadratic in y solve for y we get y=-(1+(sqrt(1-(4x^3)))) ignoring other root because for that if we substitute back we get x=1 as solution which is not possible as roots are complex hence now substitute this y in second equation and square it and simplify we get 9th degree equation in x product of roots is the constant term which is 11 hence answer

Notice that the system of equations can be rearranged to get $\begin{cases}y^2+y+1 &= 2-x^3\\y^3 - 1 &= 1-x^2\end{cases}$ Since from the second equation $y^3-1=(y-1)(y^2+y+1)=(y-1)(2-x^3)=1-x^2,$ we have that $y-1=\dfrac{1-x^2}{2-x^3}$ and $y=\dfrac{3-x^2-x^3}{2-x^3}$ . Note that it is possible to divide by $2-x^3$ since if $2-x^3=0$ , then $1-x^2$ (the right-hand side of the equation) must be equal to $0$ as well, and there is no number $x$ such that both $x^3=2$ and $x^2=1$ . Substituting this into the original second equation gives [\begin{align*}x^2+\left(\dfrac{3-x^2-x^3}{2-x^3}\right)^3&=2\x^2(2-x^3)^3+(3-x^2-x^3)^3&=2(2-x^3)^3\ x^2-2)(2-x^3)^3+(3-x^2-x^3)^3&=0\end{aligned} By Vieta's Formulas, the absolute value of the product of the roots of any polynomial equation is equal to the absolute value of the result when the constant term of the polynomial is divided by its leading coefficient. First, it may be deduced that the leading coefficient of the LHS is $-1$ . This is because the left product of polynomials has the only possible $x^{11}$ term on the LHS, and multiplying it out shows that the coefficient on this term is $-1$ . Next, it can be deduced that since the constant term of $(x^2-2)(2-x^3)^3$ is $-2\cdot 2^3=-16$ and the constant term of $(3-x^2-x^3)^3$ is $3^3=27$ , the constant term of the entire polynomial is $-16+27=11$ . Therefore, by representing the roots of the original system as $x_1,x_2,\ldots,x_{11}$ we have that $|x_1x_2x_3\cdots x_{11}=\left|\dfrac{11}{-1}\right|=11$ and we are done.