What if the roots of this equation are in geometric progression?

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Since we know the roots are in geometric progression, and the leading coefficient is \(1\), we have \[x^3 - 5x^2 + qx - 8 = (x-c)(x-cr)(x-cr^2),\] so \[x^3 - 5x^2 + qx - 8 = x^3 - c(1 + r + r^2) x^2 + c^2(r + r^2 + r^3) x - c^3 r^3,\] and this is true for all values of \(x\).

Comparing coefficients of the different powers of \(x\) now tells us that

1. \(c^3 r^3 = 8\),
2. \(c(1+r+r^2) = 5\), and finally that
3. \(q = c^2r(1+r+r^2)\).

But now notice that \[q = c(1+r+r^2) \times cr = 5 \times 8^{1/3} = 10,\] as desired.

Using the same approach as above, we find that

1. \(\alpha + \beta + \gamma = 1\),
2. \(\alpha\beta + \beta\gamma + \gamma\alpha = 4\),
3. \(\alpha\beta\gamma = -7\).

Suppose the polynomial we need is of the form \[x^3 + A x^2 + B x + C = (x-\beta-\gamma)(x-\gamma - \alpha)(x-\alpha-\beta).\]

Expanding gives

1. \(-A = (\beta+\gamma)+(\gamma+\alpha)+(\alpha+\beta),\)
2. \(B = (\beta+\gamma)(\gamma + \alpha)+(\beta+\gamma)(\alpha+\beta)+(\gamma + \alpha)(\alpha+\beta),\)
3. \(-C = (\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)\).

Consider first \(A\). We see that \[-A = (2\alpha + 2\beta + 2\gamma) = 2(\alpha+\beta+\gamma) = 2,\] so \(A = -2\).

Next consider \[\begin{align*} B &= (\beta+\gamma)(\gamma + \alpha)+(\beta+\gamma)(\alpha+\beta)+(\gamma + \alpha)(\alpha+\beta)\\ &= \alpha^2 + \beta^2 + \gamma^2 + 3\alpha\beta+3\beta\gamma+3\gamma\alpha\\ &=\alpha^2 + \beta^2 + \gamma^2 + 2\alpha\beta+2\beta\gamma+2\gamma\alpha + (\alpha\beta+\beta\gamma+\gamma\alpha)\\ &=(\alpha+\beta+\gamma)^2+4\\ &=5. \end{align*}\] Before we tackle \(C\), let’s note that \[\begin{align*} 1 \times 4 &= (\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha)\\ &= \alpha^2\beta+\alpha^2 \gamma+\beta^2\alpha+ \beta^2\gamma + \gamma^2\alpha+\gamma^2\beta+3\alpha\beta\gamma\\ &= \alpha^2\beta+\alpha^2 \gamma+\beta^2\alpha+ \beta^2\gamma + \gamma^2\alpha+\gamma^2\beta-21\\ &\implies \alpha^2\beta+\alpha^2 \gamma+\beta^2\alpha+ \beta^2\gamma + \gamma^2\alpha+\gamma^2\beta =25.\\ \end{align*}\] Now we have \[\begin{align*} -C &= (\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)\\ &= \alpha^2\beta+\alpha^2 \gamma+\beta^2\alpha+ \beta^2\gamma + \gamma^2\alpha+\gamma^2\beta + 2 \alpha\beta\gamma\\ &= 25 -14 = 11, \end{align*}\]

and so \(C = -11.\)

Therefore the required equation is \[x^3 - 2x^2 + 5x - 11=0.\]