#### Question

Let x_1 and x_2 be the roots of the equation x^2 − (a + d)x + (ad − bc) = 0. Show that x^3_1 and x^3_2 are the roots of the equation

y^2 − (a^3 + d^3 + 3abc + 3bcd)y + (ad − bc)^3 = 0

Question

Let x_1 and x_2 be the roots of the equation x^2 − (a + d)x + (ad − bc) = 0. Show that x^3_1 and x^3_2 are the roots of the equation

y^2 − (a^3 + d^3 + 3abc + 3bcd)y + (ad − bc)^3 = 0

Since x_1 and x_2 be the roots of the equation x^2 − (a + d)x + (ad − bc) = 0, according to Vieta's formula,

x_1+x_2 = a+d

(1)

and

x_1x_2 = ad − bc

(2)

x^3_1+x^3_2 = (x_1+x_2)^3-3x_1x_2(x_1+x_2)

=(a+d)^3-3(ad-bc)(a+d)

Therefore,

x^3_1+x^3_2=a^3 + d^3 + 3abc + 3bcd

(3)

On the other hand, cubing the equation (3) gives

x^3_1x^3_2 = (ad − bc)^3

(4)

Substituting (3) and (4) to the equation

y^2 − (a^3 + d^3 + 3abc + 3bcd)y + (ad − bc)^3 = 0

We get

y^2 − (x^3_1+x^3_2)y +x^3_1x^3_2 = 0

Apply Vieta's formula for a quadratic equation, x^3_1 and x^3_2 are the roots of the equation.