Shouldn't the first equation have a -3? I'm not saying it's a problem, I'm saying it for symmetry

MiltonMath12 wrote: Shouldn't the first equation have a -3? I'm not saying it's a problem, I'm saying it for symmetry We could of course have any values we like on the right hand sides of the equations. I think that the ones in the question were chosen so that $x$, $y$, and $z$ are real. Though they're still not "nice" values, so I don't know if it really gains us anything.

DylanN wrote: If $x$, $y$, $z$ are real numbers satisfying \begin{align*} (x + 1)(y + 1)(z + 1) & = 3 \\ (x + 2)(y + 2)(z + 2) & = -2 \\ (x + 3)(y + 3)(z + 3) & = -1, \end{align*}find the value of $$ (x + 20)(y + 20)(z + 20). $$ Let's define $\sigma_j$ as elementary symmetric functions of polynomials in one variable . Let's associate $(x,y,z)$ to a polynomial in $\mathbb{R}[t]$, then the system is equivalent to $$\begin{aligned} \sigma_3 +\sigma_2 +\sigma_1 =2 \\ \sigma_3 +2\sigma_2 +4\sigma_1 = -10 \\ \sigma_3 +3\sigma_2 +9\sigma_1=-28 \end{aligned}$$Solving this we get $(\sigma_1 ,\sigma_2, \sigma_3)=(-3,-3,8)$ Using these , we get $$(x+20)(y+20)(z+20)=\sigma_3 +20 \sigma_2 +400\sigma_1+8000=6748$$@below I saw that quite late, so the desired edits have been made...Thanks anyway..

homotopygroup wrote: Using these , we get $$(x+20)(y+20)(z+20)=\sigma_3 +20 \sigma_2 +400\sigma_1=-1252$$ There is a small mistake in your solution. We should have that $$ (x + 20)(y + 20)(z + 20) = \sigma_3 +20 \sigma_2 +400\sigma_1 \color{red}{+ 8000}. $$

DylanN wrote: If $x$, $y$, $z$ are real numbers satisfying \begin{align*} (x + 1)(y + 1)(z + 1) & = 3 \\ (x + 2)(y + 2)(z + 2) & = -2 \\ (x + 3)(y + 3)(z + 3) & = -1, \end{align*}find the value of $$ (x + 20)(y + 20)(z + 20). $$ Let $x+2=a,y+2=b,z+2=c,$ so $$abc=-2.$$$$(a+1)(b+1)(c+1)=-1\implies \sum_{cyc}a+\sum_{cyc}ab=0,$$$$(a-1)(b-1)(c-1)=3\implies \sum_{cyc}a-\sum_{cyc}ab=6,$$$$\sum_{cyc}a=3,\sum_{cyc}ab=-3.$$$$ (x + 20)(y + 20)(z + 20)=(a+18)(b+18)(c+18)=abc+18\sum_{cyc}ab+{18}^2\sum_{cyc}a+{18}^3=6748. $$

DylanN wrote: If $x$, $y$, $z$ are real numbers satisfying \begin{align*} (x + 1)(y + 1)(z + 1) & = 3 \\ (x + 2)(y + 2)(z + 2) & = -2 \\ (x + 3)(y + 3)(z + 3) & = -1, \end{align*}find the value of $$ (x + 20)(y + 20)(z + 20). $$

Attachments:

Define the monic cubic polynomial $P(i)=(i+x)(i+y)(i+z)$ We have $P(i)=i^3+ai^2+bi+c$ Using the given equations, we get the following system $$a+b+c=2$$$$4a+2b+c=-10$$$$9a+3b+c=-28$$ $$3a+b=-12$$$$5a+b=-18$$ $$a=-3$$$$b=-3$$$$c=8$$ So $P(20)=8000-3\cdot 400-3\cdot 20+8=\boxed{6748}$