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

Uniteasy Admin · Accepted Answer

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,
$$x_1+x_2 = a+d$$n
and
$$x_1x_2 = ad − bc$$n
$$x^3_1+x^3_2 = (x_1+x_2)^3-3x_1x_2(x_1+x_2)$$
$$=(a+d)^3-3(ad-bc)(a+d)$$
Therefore,
$$x^3_1+x^3_2=a^3 + d^3 + 3abc + 3bcd$$n
On the other hand, cubing the equation (3) gives
$$x^3_1x^3_2 = (ad − bc)^3$$n
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.

Question

Answer