# Prove that if $2a^3 + 27c = 9ab,$ then the roots of $x^3 + ax^2 + bx + c = 0$ form an arithmetic sequence.

I am not sure how to begin this problem. Can someone help me?

I have a hint: Let $$y = x + \frac{a}{3}$$ and rewrite $$x^3 + ax^2 + bx + c = 0$$ in terms of $$y.$$

How do I do this?

• Welcome to the site. If you have any doubts about formating check math.meta.stackexchange.com/questions/5020/… or ask in mathematics meta. Mar 20, 2020 at 16:22
• Note that the effect of the "hint" translates the roots of the original polynomial by $-a/3$, which preserves the arithmetic progression of three roots, if originally present. So the hint makes sense; you should report your results in rewriting the polynomial "in terms of $y$." Mar 20, 2020 at 16:22
• You have three roots in arithmetic progression, therefore their sum is equal to three times the middle root. That sum by Vieta's Formula is $-a$ and therefore the middle root is... . Plug that proposed middle root into the equation and verify that the polynomial has a zero there iff the claimed relation between the coefficients holds true. Mar 20, 2020 at 16:22
• The hint is not that much, you get a polynomial without the $x^2$ coefficient. A way to do is just put an identification system as $(x-d)(x-d-r)(x-d-2r)=x^3+ax^2+bx+c$ you obtain a system of three equations and you get all the roots in terms of $a,b,c$ with the necessary condition you want. Mar 20, 2020 at 16:27
• @ToniMhax the $c$ on the left is not the same as the $c$ on the right. Please use different letter Mar 20, 2020 at 16:30

Well, do exactly what the hint tells you!

$$(y - \frac{a}{3})^3 + a(y-\frac{a}{3})^2 + b(y-\frac{a}{3}) + c = 0$$

$$\frac{1}{27}(2a^3 - 9a^2y - 9ab + 27by +27c + 27y^3)$$

$$\frac{1}{27}( - 9a^2y + 27by + 27y^3)$$ $$\frac{y}{27}(27y^2 - 9a^2 + 27b)$$

Now can you see what the roots are? How does this relate to the roots of your original polynomial?

The roots satisfy

$$x_1+x_2+x_3=-a\tag 1$$

Write the given cubic equation in its depressed form with $$t=x+\frac a3$$,

$$t^3+(b-\frac {a^3}3)t+\frac{2a^2+27c-9ab}{27}=0$$

which, from the given condition, reduces to,

$$t^3+(b-\frac {a^3}3)t=0$$

and one of the roots is $$t_1=0$$, or, $$x_1=-\frac a3$$. From (1), we have

$$x_2+x_3=-a-x_1=-\frac23a=2x_1$$

hence, an arithmetic sequence.

Isolating $$c$$ in what we are given, and then plugging it into the polynomial, we have, $$x^3+ax^2+bx+c=x^3+ax^2+bx+\frac{1}{27}(9ab-2a^3)=$$ $$=\frac{1}{27}(3x+a)(9x^2+6ax+9b-2a^2),$$ which says $$x_1=-\frac{a}{3}$$ and $$x_2+x_3=-\frac{6a}{9}=-\frac{2a}{3}=2x_1$$ and since $$x_2-x_1=x_1-x_3,$$ we are done!

WLOG let us assume the roots to be $$p-q,p,p+q$$

$$-a=p+q+p+p-q\iff p=-\frac a3$$

$$c=p(p^2-q^2)=-\frac{a}3\left(\frac{a^2}{9}-q^2\right)\implies q^2=?$$

$$b=p(p+q)+(p-q)(p+q)+p(p-q)=p^2-q^2+2p^2=3\left(-\frac{a}3\right)^2-q^2\implies q^2=?$$

Try to compare the two values of $$q^2$$ to eliminate $$q$$.

Suppose that $$u,v,w$$ are the three roots of $$x^3+ax^2+bx+c$$. Then, because $$(x-u)(x-v)(x-w)=x^3-(\overbrace{u+v+w}^{-a})x^2+(\overbrace{uv+vw+wu}^b)x-\overbrace{\ \ u\ v\ w\ \ }^{-c}$$ we have $$2(\overbrace{u+v+w}^{-a})^3+27\overbrace{\ \ u\ \ v\ \ w\ \ }^{-c}-9(\overbrace{u+v+w}^{-a})(\overbrace{uv+vw+wu}^b)\\ =\underbrace{(u+v-2w)}_{\substack{\text{0 if w is midway}\\\text{between u and v}}}\underbrace{(2u-v-w)}_{\substack{\text{0 if u is midway}\\\text{between v and v}}}\underbrace{(u-2v+w)}_{\substack{\text{0 if v is midway}\\\text{between u and w}}}$$