Answer to Question #245414 in Algebra for Oliver

Question

Equation x^3 + 27x^2 + p = 0 has three different real solutions that form an arithmetic progression. Find the smallest value of parameter p.

Answer 1

Suppose $a, a+d, a+2d$ are the roots of the cubic equation

$$x^3 + 27x^2 + p = 0$$

Then

$$x^3 + 27x^2 + p =(x-a)(x-(a+d))(x-(a+2d))$$

We have

$$a+a+d+a+2d=-\dfrac{27}{1}$$

$$a(a+d)+a(a+2d)+(a+d)(a+2d)=\dfrac{0}{1}$$

$$a(a+d)(a+2d)=-\dfrac{p}{1}$$

$$d=-9-a$$

$$a(-9)+a(-a-18)+(-9)(-a-18)=0$$

$$-9a-a^2-18a+9a+162=0$$

$$a^2+18a-162=0$$

$$D=(18)^2-4(1)(-162)=972$$

$$a=\dfrac{-18\pm\sqrt{972}}{2(1)}=-9\pm9\sqrt{3}$$

$$a=-9-9\sqrt{3}, d=9\sqrt{3}$$

$$p=-(-9-9\sqrt{3})(-9)(-9+9\sqrt{3})=-1458$$

$$a=-9+9\sqrt{3}, d=-9\sqrt{3}$$

$$p=-(-9+9\sqrt{3})(-9)(-9-9\sqrt{3})=-1458$$

$p=-1458$