Question 1176146
<pre>
{{{x^3 + ax^2 + bx + c = 0}}} 

Suppose the roots (which form an arithmetic sequence) are p-d, p, and p+d.
Then

The sum of the roots is -a
 
{{{(p-d) + p + (p+d) = -a}}}

{{{p - d + p + p + d = -a}}}

{{{3p = -a}}}

{{{a=-3p}}}

The sum of the products of pairs of roots is b

{{{(p-d)p + (p-d)(p+d) + p(p+d) = b}}}

{{{p^2 - dp + p^2 - d^2 + p^2 + pd = b}}}

{{{3p^2 = b}}}

The product of the roots is -c

{{{(p-d)(p)(p+1) = -c}}}

{{{p(p-d)(p+d) = -c}}}

{{{p(p^2-d^2) = -c}}}

{{{p^3-pd^2 = -c}}}
{{{c = pd^2-p^3}}}

So we have solved a, b, and c in terms of p and d

Substitute in the original equation:

{{{p^3 + (-3p)p^2 + (3p^2)p + (p*d^2-p^3) = 0}}}

{{{p^3-3p^3+3p^3+pd^2-p^3=0}}}

{{{pd^2=0}}}

So either p=0 or d=0

If p=0, then the roots are -d, 0, and d

Then the sum of the roots = -d+0+d = 0 = -a, so a=0

The sum of the products of pairs of roots = (-d)(0)+(-d)(d)+(0)(d)=-d<sup>2</sup>, so b=-d<sup>2</sup>

Then the product of the root is (-d)(0)(d) = 0, so c=0

That means the original equation, in this case, was really:

{{{x^3+0x^2-d^2x+0=0}}} or

{{{x^3-d^2x=0}}}

So we see if {{{2a^3 + 27c = 9ab}}} holds true in this case:

{{{matrix(1,3,2a^3+27c,"?=?",9ab)}}}

{{{matrix(1,3,2(0)^3+27(0),"?=?",9(0)(0 +- d))}}}

{{{matrix(1,3,0,"=",0)}}}

So yes it does hold when p=0

Now we see what happens when d=0.

Then the roots are p-0, p, and p+0, or p, p, and p. 

So the three roots are all equal.

The sum of the roots is 3p, so a=-3p

The sum of the products of pairs of roots = (p)(p)+(p)(p)+(p)
(p)=3p<sup>2</sup>, so b=3p<sup>2</sup>

Then the product of the roots is (p)(p)(p) = p<sup>3</sup>, so c=-p<sup>3</sup>

That means the original equation, in this case, was really:

{{{x^3+ax^2+b^2x+c=0}}} or

{{{x^3-3px^2+3p^2x-p^3=0}}} which is just {{{(x-p)^3=0}}}

So we see if {{{2a^3 + 27c = 9ab}}} holds true in this case as well:

{{{matrix(1,3,2a^3+27c,"?=?",9ab)}}}

{{{matrix(1,3,2(-3p)^3+27(-p^3),"?=?",9(-3p)(3p^2))}}}

{{{matrix(1,3,2(-27p^3)-27p^3,"?=?",-81p^3)}}}

{{{matrix(1,3,-54p^3-27p^3,"?=?",-81p^3)}}}

{{{matrix(1,3,-81p^3,"?=?",-81p^3)}}}

So yes it does hold true in this case too.

The (a) part of the problem is proved.

If I find time I'll do (b) as well.
 
Edwin</pre>