Here ∝ = 1, β = 1 and ૪ = -2
∴ The required cubic equation is
x³-(1+1-2)x²(1-2-2)x-(1)(1)(-2) = 0
x³- 0x²-3x+2 = 0
x³-3x-2 = 0

Let p(x) = x⁹ + 9x⁷ + 7x⁵ + 5x³ + 3x
p(x) has no sign change.
p(-x)⇒ (-x)⁹ + 9(-x)⁷ + 7(-x)⁵ + 5(-x)³ + 3(-x)
= -x⁹ - 9x⁷ - 7x⁵- 5x³ -3x
p(-x) also has no sign change
∴ p(x) has no positive and no negative root.

Here ∝ = 2, β = -2 and ૪ = 4
x³-(2-2+4)x²+(- 4 - 8 + 8x)x-(2)(-2)(4) = 0
⇒x³- 4x²- 4x+16 = 0

Let P(x) be the polynomial under consideration.
The number of sign changes for P(x) and P(−x) are 2 and 1 respectively. Hence it has at most two positive roots and at most one negative root. Since the difference between number of sign changes in coefficients of P(−x) and the number of negative roots is even, we cannot have zero negative roots. So the number of negative roots is 1. Since the difference between number of sign changes in coefficient of P(x) and the number of positive roots must be even, we must have either zero or two positive roots. But as the sum of the coefficients is zero, 1 is a root. Thus we must have two and only two positive roots Obviously the other two roots are imaginary numbers.

x⁸- 3x + 1 = 0
Here a_n = 1, a_o = 1
If \(\frac{p}{q}\) is a root of the polynomial, then as 
(p, q) = 1p is a factor of a_o = 1 and q is a factor of a_n = 1
Since 1 has no factors, the given equation has no rational roots.