If roots of $x^3 + 5x^2 - 7x - 1 = 0$ are $\alpha, \beta, \gamma$, then the equation whose roots are $\alpha \beta, \beta \gamma, \gamma \alpha$, is
- $x^3 - 7x^2 + 5x + 1 = 0$
- $x^3 + 7x^2 - 5x - 1 = 0$
- $x^3 + 5x^2 + 7x + 1 = 0$
- $none\ of\ these$
Given x^3 + 5x^2 - 7x - 1 = 0, the roots are a, b, c. We have a+b+c = -5, ab+bc+ca = -7, and abc = 1. We want an equation with roots ab, bc, ca. Let y = ab = 1/c, y = bc = 1/a, y = ca = 1/b. Substituting c = 1/y into the original equation: (1/y)^3 + 5(1/y)^2 - 7(1/y) - 1 = 0. Multiplying by -y^3 gives y^3 + 7y^2 - 5y - 1 = 0.
Let the new roots be $y = \alpha\beta$, $z = \beta\gamma$, and $w = \gamma\alpha$. Using Vieta's formulas for the original cubic $x^3 + 5x^2 - 7x - 1 = 0$, we have $\alpha+\beta+\gamma = -5$, $\alpha\beta+\beta\gamma+\gamma\alpha = -7$, and $\alpha\beta\gamma = 1$. The sum of the new roots is $y+z+w = \alpha\beta+\beta\gamma+\gamma\alpha = -7$, their pairwise sum is $yz+yw+zw = \alpha\beta^2\gamma+\beta\gamma^2\alpha+\gamma\alpha^2\beta = \alpha\beta\gamma(\alpha+\beta+\gamma) = (1)(-5) = -5$, and their product is $yzw = \alpha^2\beta^2\gamma^2 = 1$. The required monic polynomial is $x^3 - (\text{sum})x^2 + (\text{pairwise sum})x - \text{product} = 0$, which becomes $x^3 + 7x^2 - 5x - 1 = 0$.