MHB How Do You Solve POTW #456 for (x+20)(y+20)(z+20)?

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The problem involves finding the value of (x+20)(y+20)(z+20) given three equations involving x, y, and z. The equations are (x+1)(y+1)(z+1)=3, (x+2)(y+2)(z+2)=-2, and (x+3)(y+3)(z+3)=-1. Members Theia and kaliprasad provided correct solutions to the problem. Theia's solution details the steps taken to derive the final value. The discussion highlights the collaborative effort in solving complex polynomial equations.
anemone
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Here is this week's POTW:

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If $x,\,y$ and $z$ are real numbers satisfying

$(x+1)(y+1)(z+1)=3\\(x+2)(y+2)(z+2)=-2\\(x+3)(y+3)(z+3)=-1$

find the value of $(x+20)(y+20)(z+20)$.

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Congratulations to the following members for their correct solution, which you can find below:

1. Theia
2. kaliprasad

Solution from Theia:
We have

$$\begin{cases}(x + 1)(y + 1)(z + 1) = 3 \\ (x + 2)(y + 2)(z + 2) = -2 \\ (x + 3)(y + 3)(z + 3) = -1 \end{cases}$$

and we want to calculate \(L = (x + 20)(y + 20)(z + 20).\)
Let's write

$$\begin{cases}x = a - 1 \\ y = b - 1 \\ z = c - 1.\end{cases}$$

After substituting to the original equation and expanding we have

$$\begin{cases}abc = 3 \\ abc + ab + ac + bc + a + b + c + 1 = -2 \\ abc + 2(ab + ac + bc) + 4(a + b + c) + 8 = -1.\end{cases}$$

Now substitute \(abc = 3\) into two other equations and write

$$\begin{cases}ab + ac + bc = q \\ a + b + c = r.\end{cases}$$

Hence we have obtained simultaneous equations

$$\begin{cases}q + r = -6 \\ 2q + 4r = -12, \end{cases}$$

whose solution is \(q = -6, r = 0.\) The expression \(L\) is in terms of \(a,b,c\):

$$\begin{align*}L &= (a + 19)(b + 19)(c + 19) \\ &= abc + 19(ab + ac + bc) + 19^2(a + b + c) + 19^3 \\ &= 3 + 19\cdot q + 19^2\cdot r + 19^3 \\ &= 3 - 6\cdot 19 + 19^2\cdot 0 + 19^3 \\ &= 6748.\end{align*}$$

Hence \(L = (x + 20)(y + 20)(z + 20) = 6748.\)
 
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