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

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

So far, I let the roots of $$x^3+ax^2+bx+c=0$$ be $$r_1, r_2,$$ and $$r_3$$. $$r_1=r_2-d$$ and $$r_3=r_2+d$$ because they form an arithmetic sequence with $$d$$ being the difference. the sum of the roots is $$-a$$. So, $$r_2=-a/3$$. We can let the product of the roots be $$-c$$. So, $$(r_2-d)(r_2)(r_2+d)=-c$$. Plugging in $$r_2=-a/3$$ we get $$(-a/3-d)(-a/3)(-a/3+d)$$. How do I continue with this method?

EDIT: I used hamam_abdallah's hint to get $$\frac{-a^3}{27} + \frac{ad^2}{3} = -c$$ what do i do after applying vieta's formulas?

Put $$a=3A$$ The equality to prove $$2a^3+27c=9ab$$ becomes

$$\boxed{2A^3+c=Ab}$$

As you said $$r_2=\frac{-a}{3}= -A$$ is a root of $$x^3+3Ax^2+bx+c=0$$ then

$$(-A)^3+3A(-A)^2+b(-A)+c=0$$

$$\iff \; -A^3+3A^3+c=Ab$$ $$\iff \; \boxed{2A^3+c=Ab}$$ Done.

Hint:

Use Vieta formulas: $$r_1+r_2+r_3 =-a$$ $$r_1r_2+r_2r_3+r_3r_1=b$$ and $$r_1r_2r_3 =-c$$

Hint

Finally note that $$(-a/3-d)(-a/3+d) +(-a/3)(-a/3+d)+(-a/3-d)(-a/3)=b$$ gives $$(-a/3-d)(-a/3+d)= b-(-a/3)(-a/3+d)-(-a/3-d)(-a/3)=b+ (a/3)(-a/3+d-a/3-d)$$

Thus $$(-a/3-d)(-a/3+d)=b- (2a^2/9)$$

Therefore $$-c=(-a/3-d)(-a/3)(-a/3+d)=(-a/3)\left( b- (2a^2/9) \right)$$

Let's do this thing

$$(r+d)^3 +a(r+d)^2 +b(r+d) + c =(r-d)^2 + a(r-d)^2 + b(r-d) + c = r^3 + ar^2 +br + c = 0$$ (read the hidden for the insight as to how I thought to do the next step... or just go on to the next step)

Insight: Now the standard trick is to note that the coefficients of the even powers of $$d$$ in $$(r+d)^3 +a(r+d)^2 +b(r+d) + c$$ and in $$(r-d)^2 + a(r-d)^2 + b(r-d) + c$$ are the same and the coefficients of the odd powers of $$d$$ are equal be opposite signs. Yet they both sum to the same value. So the sum of the coefficients of the odd powers add to $$0$$. That is:

so $$\frac {[(r+d)^3 +a(r+d)^2 +b(r+d) + c]-[(r-d)^2 + a(r-d)^2 + b(r-d) + c]}2=0$$ so

$$3r^2d + d^3 + 2ard + bd = 0$$

Insight: Now we have $$(r+d)^3 +a(r+d)^2 +b(r+d) + c = r^3 + ar^2 + br +c$$ and this is a little less of a standard trick but that means the sum of the non-negative powers of $$d$$ add to zero. And as we know the sum of the odd powers add to zero, that means the sum of the even powers add to zero as well. That is:

And $$[(r+d)^3 +a(r+d)^2 +b(r+d) + c]-[3r^2d + d^3 + 2ard + bd]-[r^3 + ar^2 +br + c]=0$$ so

$$3rd^2 + ad^2= 0$$

If $$d$$ is the incremental of an arithmetical sequence then $$d$$, presumably does not equal $$0$$. Otherwise the "sequence" is constant. Which.... technically is an arithmetic sequence let's put a pin in that...

and if we assume $$d\ne 0$$

so $$r=-\frac a3$$.... ooookay.... I wasn't expecting that....

$$-\frac {a^3}{27} + a\frac {a^2}9 -b\frac a3 + c = \frac {2a^3}{27}-\frac {ab}3 + c =0$$ so

$$2a^3 +27c = 9ab$$.

.....

And if $$d = 0$$ then $$r+d = r=r-d$$ and $$r$$ is a triple root.

So $$x^3 + ax^2 + bx + c = (x-r)^3$$

So $$a = -3r$$ and $$b=3r^2$$ and $$c= -r^3$$. So $$2a^3 + 27c = -2*27r^3 - 27r^3 = -81r^3$$. And $$9ab= -9*3r*3r^2 =-81r^3$$.