Coefficient Cubic Roots

From the Vieta's formulas, we can find three equations:

$1) \frac{-a}{1} = a + b + c$

$2) \frac{b}{1} = ab + bc + ac$

$3) \frac{-c}{1} = abc$

Let's start by the third equation, which will divide our problem into two cases. If $\frac{-c}{1} = abc$ , we can consider $c=0$ or $c\neq0$ .

$(a) c=0$

If $c=0$ , the second equation may be simplified as $b = ab$ . Again, either $b=0$ or $b\neq0$ .

If $b\neq 0, a=1$ . Using the first equation, $-1 = 1 + b$ , whic means $b=-2$ . First answer: $(1,-2,0)$ .

If $b=0$ , the first equation tells us that $a=-a$ , so $a=0$ . Second answer: $(0,0,0)$ .

$(b) c\neq 0$

Still from the third equation, the following is found: Eq. (3): $ab=-1$ . Therefore, we have a system:

Eq. (2): $b=-1+(a+b)c$

Eq. (1): $2a + b + c = 0$

Rewriting the system with all variables in function of a, we find:

Eq. (3): $b= \frac{-1}{a}$

Eq. (1): $c=-2a+\frac{1}{a}$

Now, eq. (2): $\frac{-1}{a}=-1+(a-\frac{1}{a})(-2a+\frac{1}{a})$

Simplifying: Eq. (*): $a-1=(a^2-1)\frac{-2a^2+1}{a}$

From this last line, we have two more cases: $a=1$ or $a\neq 1$ .

If $a=1, b=\frac{-1}{1}, c=-2*1+\frac{1}{1}=-1$ , so we have a third answer: $(1,-1,-1)$ .

If $a\neq 1$ , the equation becomes:

Eq. (*): $a=(a+1)(-2a^2+1)$

Simplifying, eq. (*): $2a^3+2a^2-1=0$ .

By testing a=1 in (*) above, $2*1+2*1-1\neq 0$ , so it's not a possible root.

Let's use the Descartes' rule of signs: the number of positive roots of the polynomial is either equal to the number of sign differences between consecutive nonzero coefficients, or is less than it by an even number. Since there is only one sign difference, there is only one positive root for this polynomial (the smallest even number is 2, and $1-2<0$ , so 1 is the only possibility for the number of roots). That one positive root is not 1, as we found above, so this is a new and fourth answer.

Let's calculate the derivative of the polynomial: $3*2a^2+2*2a^1-0$

The points of local maximum and minimum values are the roots of the equation above:

$3*2a^2+2*2a^1-0=0, 6a^2+4a=0$

Roots: 0 and $\frac{-2}{3}$ . Calculating the correponding values of the function (*):

$2*0^3+2*0^2-1=-1$ and $2(\frac{-2}{3})^3+2(\frac{-2}{3})^2-1<0$ .

Since both are negative, it means the curve of the graph doesn't cross the x axis more than once, therefore there is only one real root (the positive one) for the polynomial equation (*).

Final answer: There are four possible sets for the problem

Use Vieta's to get

$a + b + c = -a (1)$

$ab + bc + ac = b (2)$

$abc = -c (3)$

From $(3)$ we get that $c(ab + 1) = 0$ . It's only logical now to break it into two cases:

Case 1: $c = 0$ .

We have that $2a = b$ and $ab = b \rightarrow b(a - 1) = 0$ . Substituting the first equation into the second, we get $2a(a - 1) = 0$ , so $a = 0, a = 1$ and $b = 1, b = -2$ . To recap, $(0, 0, 0), (1, -2, 0)$ are solutions.

Case 2: $ab = -1$ . From $(2)$ we have $bc + ac = b + 1$ , so $c = \frac{b+1}{a+b}$ . But hold on, before we do that, let's make sure that $a + b \neq 0$ . If this were the case, then $a = 1, b = -1$ from our conditions. Plugging in these values into our equations we get an answer in $c = -1$ . So, $1, -1, -1$ is a solution. Now that we found our solutions for $a+b = 0$ , let's assume $a+b \neq 0$ from now on. Going back to our equation, we substitute $a = \frac{-1}{b}$ , clear complex fractions, factor and reduce to get $c = \frac{b}{b-1}$ . In addition, doing the same with the equation $a+b+c = -a$ , we get that $c = \frac{2}{b} - b$ . Setting the two equations together, we get a cubic in $b^3 - 2b + 2 = 0$ . We need to know if there are one or three real roots in this. We can do this with a little bit of calculus.

Taking the derivative with respect to b, we want to find where the curve has a slope of zero. $\frac{dy}{db} = 3b^2 - 2 = 0 \rightarrow b = \pm \sqrt{\frac{2}{3}}$ . Plugging these values into the original equation we find that both y-values are positive, and using the second derivative test we can affirm these as local maxima and minima. Thus, the function only has one root, occurring somewhere to the left of $b = \sqrt{\frac{2}{3}}$ , giving us a total of $\boxed{4}$ answers.

If a.b.c are roots of this cubic

we have sum of roots = a+ b+ c = -a ( coefficient of x^2) b+ c + 2a =0

sum of roots two at a time = ab + bc + ca = b ( coefficient of x)

product of roots = abc = -c ( coefficient of constant) either c =0 or ab =-1

if c = 0 we get two solutions by solving equation which are (0,0,0) and ( 1,-2,0)

if c <> 0 ( c not equal to 0) and ab =-1

solving equation be get 2a^2 + b^2 + b = 2

2/b^2 + b^2 + b = 2

b^4+ b^3 -2b^2 +2 = 0

(b+1) ( b^3 - 2b+2 ) = 0

first part of the equation gives b = -1 which gives b =-1, a =1 , c =-1

second part of above equation has two imaginary roots, and other irrational root but real root.

EDITED PART for ( b^3 - 2b+2 ) = 0 Using http://www.1728.org/cubic.htm we get another real value of b = -1.76935292423863 a = 0.56519771738364 c = 0.638896919471351

hence we have four sets of solution (0,0,0) (1,-1,-1) (1,-2,0) ( 0.56519771738364,-1.76935292423863,0.638896919471351)

If a.b.c are roots of this cubic

we have sum of roots = a+ b+ c = -a ( coefficient of x^2) b+ c + 2a =0

sum of roots two at a time = ab + bc + ca = b ( coefficient of x)

product of roots = abc = -c ( coefficient of constant) either c =0 or ab =-1

if c = 0 we get two solutions by solving equation which are (0,0,0) and ( 1,-2,0)

if c <> 0 ( c not equal to 0) and ab =-1

solving equation be get 2a^2 + b^2 + b = 2

2/b^2 + b^2 + b = 2

b^4+ b^3 -2b^2 +2 = 0

(b+1) ( b^3 - 2b+2 ) = 0

first part of the equation gives b = -1 which gives b =-1, a =1 , c =-1

second part of above equation has two imaginary roots, and other irrational root but real root.

we get another real value of b = -1.76935292423863 a = 0.56519771738364 c = 0.638896919471351

hence we have four sets of solution (0,0,0) (1,-1,-1) (1,-2,0) ( 0.56519771738364,-1.76935292423863,0.638896919471351

Generally, if t, u, v are the roots of the cubic equation $x^3 + ax^2 + bx+c=0$ , then $x^3 + ax^2 + bx + c \equiv (x-t)(x-u)(x-v)$ which gives:

$t+u+v = -a, \ tu+uv+vt = b, \ tuv = -c$

In our case, t=a, u=b, v=c. So the third equation gives $abc=-c$ and either c=0 or ab=-1. Let's consider these two cases.

Case 1: c=0.

The equation $x^2 + ax + b = 0$ thus has roots a and b, so we have ab = b, which gives b=0 or a=1. On the other hand a+b = -a, so this gives us (a,b) = (0,0) and (1,-2) respectively.

Case 2: ab=-1.

Thus a= -1/b. Thus $b = ab + bc+ca = -1 + bc -c/b$ so $b+1 = \frac c b(b^2-1)$ . Either b+1 = 0 or $1 = \frac c b (b-1)$ . Now the former gives us a = -1/b = 1 and thus c = -(2a+b) = -1. This gives (a,b,c) = (1,-1,-1).

The latter gives us $c(b-1) = b$ . Together with $2a+b+c=0\implies -\frac 2 b + b + c = 0$ , substitution gives us a cubic equation in b:

$b^3 - 2b + 2 = 0$

It's easy to see that this has only one real root (e.g. by plotting its graph). Corresponding to this b, one also checks that $(a,b,c) = (-\frac 1 b, b, \frac b{b-1})$ is a legitimate solution.

So altogether we get 4 solutions.

Let a,b,c equation's roots . Using The Vieti's Formule, we have:

(i) a+b+c = - a (ii) ab + ac + bc = b (iii) abc = - c

Using the equation (iii): c(ab+1) = 0

If c = 0 , so, by the second equation : b(a-1) = 0

if b =0 then, by the first equation, a = 0. If b isn't 0 then a=1 and, by the first equation, b = -2

Thus, we have two triples : (a,b,c)= { (0,0,0) ; (1,-2,0) }

Now, if c isn't 0, we can rewrite the three before equations:

b+c+2a =0 ac + bc - b = 1 ab = -1

Solving the system of equations, we find the following equation in the variable b:

b^4 + b^3 - 2b^2 + 2 = 0 (b+1)(b^3-2b+2)=0

If b= -1 then a=1 and c= -1 , with (a,b,c)=(1,-1,-1)

Now, we need to analyze the polynomial : b^3 - 2b + 2

By the Descartes' Rule of Signs, this polynomial has just one negative root. Therefore, there is other triple (a,b,c).

Finally, for realize that b^3-2b+2 just has one real root, we can to analyze the polynomial's graphic.

Then, we have four triples (a,b,c)

If a, b, and c are the roots of the equation x^3 + ax^2 + bx + c = 0, then the left side of the equation must factor as (x - a)(x - b)(x - c) = 0.

Multiplying out (x - a)(x - b)(x - c) = 0 gives x^3 - (a + b + c)x^2 + (ab + ac + bc)x - abc.

Thus, in order for x^3 + ax^2 + bx + c = 0 to have a, b, and c as its solutions, we must have

x^3 + ax^2 + bx + c = x^3 - (a + b + c)x^2 + (ab + ac + bc)x - abc.

Two polynomials are equal just when their coefficients are equal, which gives the equations

a = -(a + b + c) b = ab + ac + bc c = -abc.

I will solve these equations in three cases: 1) When c = 0 2) When c is not 0, and a = 1 3) When c is not 0, and a is not 1

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CASE I. Suppose that c = 0. Then the third equation is true, and the first two equations become

a = -a - b b = ab

Solving the first equation for b gives b = -2a. Plugging this in to the second equation, we obtain

-2a = a * (-2a)

which can be rearranged to form

2a^2 - 2a = 0

Factoring gives 2a(a - 1) = 0, which means a = 0 or a = 1.

Since b = -2a, then when a = 0, we get that b = 0; and when a = 1, we get that b = -2.

So (0, 0, 0) and (1, -2, 0) are two solutions.

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CASE II. Suppose that c is nonzero, and a = 1.

Since c is not zero, then we may divide both sides of the equation c = -abc by c, giving

1 = -ab. Since a = 1, then this implies b = -1.

If a = 1 and b = -1, then from the first equation, we have

1 = -(1 + (-1) + c),

which means that c = -1.

It turns out that a = 1, b = -1, c = -1 also makes the second equation true, so (1, -1, -1) is a third solution.

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CASE III. Now suppose that c is not 0, and a is not 1.

Then we may divide both sides of the third equation by c, giving 1 = -ab, or ab = -1.

Note that this means that neither a nor b can be zero.

Dividing both sides by a gives b = -1/a.

Our three equations are now

a = -(a + b + c) b = ab + ac + bc b = -1/a

Plugging in -1/a for b in the first two equations gives

a = -(a - 1/a + c) -1/a = a(-1/a) + ac + (-1/a)c

Simplifying gives

a = -a + 1/a - c -1/a = -1 + ac - c/a

Multiplying both equations by a gives

a^2 = -a^2 + 1 - ac -1 = -a + a^{2}c - c

We can now solve the second equation for c. Adding a to each side gives

a - 1 = a^{2}c - c

Factoring out c on the right gives

a - 1 = c(a^2 - 1)

Now divide by a^2 - 1 (this a legal move since we know that a is neither 0 or 1, so a^2 - 1 is nonzero) gives

(a - 1) / (a^2 - 1) = c (a - 1) / [(a - 1)(a + 1)] = c 1 / (a + 1) = c.

Plugging this into the first equation gives

a^2 = -a^2 + 1 - a / (a + 1).

Multiplying both sides by (a + 1) gives

a^2(a + 1) = -a^2(a + 1) + (a + 1) - a

which simplifies as

a^3 + a^2 = -a^3 - a^2 + a + 1 - a

Bringing everything to one side gives

2a^3 + 2a^2 - 1 = 0

Factoring gives (approximately)

(2a - 1.1303954)(a^2 + 1.56520a + 0.88465) = 0

Since a^2 + 1.56520a + 0.88465 = 0 has only complex solutions, then we must have 2a - 1.1303954 = 0; that is, a = 0.56520.

If a = 0.56520, then since b = -1/a, then b = -1.76929.

Since c = 1 / (1 + a), then c = 0.63890..

Thus, (0.56520, -1.76929, 0.63890) is a fourth solution.

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Since the three cases cover all possibilities, then there are only 4 such triples (a, b, c) with the desired property

By using Vieta's formula, we get three equations in three unknowns:

$(1) a + b + c = -a$

$(2) ab + bc + ca = b$

$(3) abc = -c$

From $(3)$ , we have $c(ab + 1) = 0$ , which gives either $c = 0$ or $ab = -1$ .

CASE 1: $c = 0$

Subsituting this into $(1)$ and $(2)$ , we have $2a + b = 0$ and $ab - b = 0$ . The latter equation gives $a = 1$ or $b = 0$ , which, when substituted into the former equation, will give $b = -2$ and $a = 0$ , respectively. Since we had found two triplets of real numbers, we count both solutions.

CASE 2: $ab = -1$

Substituting this value into $(2)$ and factoring out $c$ , we have

$(4) c (b + a) = b + 1$

From $(1)$ , we have $c = -b - 2a$ . We can also plug in $a = \frac{-1} {b}$ , implying that $b \neq{0}$ . With all these substitutions, we can rewrite $(4)$ , all fractions cleared, as:

$(5) b^4 + b^3 - 2b^2 + 2 = 0$

which can be factored into $(b + 1) (b^3 - 2b + 2) = 0$ . The third triplet comes from the solution $b = -1$ . As with the cubic polynomial, we can use Descartes' rule of signs, since it is already written in descending powers of $b$ . Let $f(b) = b^3 - 2b + 2$ . Since $f(b)$ has two variations in sign, $f$ must have two or zero positive real roots. Also, since $f(-b) = -b^3 + 2b + 2$ has one variation in sign, $f$ can have at most one negative real root.

By the Rational Zeros Theorem, we have four possible candidates for the zeros of $f$ : $-2, -1, 1,$ and $2$ . All of them do not satisfy $f(b) = 0$ . From the discriminant of the polynomial, which is $-76$ , it can be concluded that $f(b)$ has exactly one negative real root. Therefore, we have counted exactly $\boxed{4}$ real triplets $(a,b,c)$ which satisfy the problem.

Let f(x)=x^3+ax^2+bx+c. Since a,b,c should be the roots of equation hence sum of roots = a+b+c = -a => 2a+b+c=0 ....... (1)

sum of product of two roots at a time = ab+bc+ca = b => ab+bc+ca-b = 0 .......(2)

product of roots = abc = -c => ab = -1 (c not equals to 0)

now ,

CASE I :- If c = 0 from equation (1) 2a+b = 0 ........(3) and from equation (2) ab-b = 0 .......(4) solving equation (3) and (4) we get a = 1 and b = -2 . hence the first solution is (1,-2,0).

CASE II :- If c is not equal to 0 there are three equations 2a+b+c = 0 , ab+bc+ca-b = 0, ab = -1 substituting the value of b = -1/a in equations (1) and (2) we get c = (1-2a^2)/a from equation (1) and a^c-a-c+1 = 0 from equation (2) solving these two we get two different values of 'a' hence we get two values of 'b' and 'c' also . hence the sets are (1,-1,-1) and (0.5651977194,-1.769292354,0.6388969194)(approx. values) and the last set is if all the three values are 0 that means the last set is (0,0,0).

Therefore there are four sets satisfying the conditions of polynomial f(x) . 1. (1,-2,0) 2. (1,-1,-1) 3. (0.5651977194,-1.769292354,0.6388969194) 4. (0,0,0)

According to Vieta's formulas, the given polynomial P(x) = x^3 + ax^2 + bx +c can be expressed as

P(x) = x^3 + (-a-b-c)x^2 + (ab+bc+ac)x -abc

By comparing the coefficients in these two equations, we get
-a-b-c = a
ab+bc+ac = b
-abc = c

• When none of the coefficients a,b and c is zero, there is only one solution (a,b,c) = (1,-1,-1)

• When at least one coefficient, a,b or c is zero, zero is one of the polynomial roots.

  P(0) = 0^3 + a 0^2 + b 0 + c = 0
  Therefore, c must be zero.
  

• When a=0 and c=0,

  P(b) = b^3 + b^2 =0 ,
  which implies that b is also zero, (a,b,c) = (0,0,0).
  

• When b=0 and c=0,

P(a) = 2a^3 = 0, which means a is also zero.

• When only c is zero,

  P(x) = x^3 + ax^2 + b^x

  P(a) = 2a^3 + ab

  P(b) = b^3 + (a+1)b^2

By solving this system of equations, we get two triples, $(1,-2,0)$ and $( \frac{1}{2} , - \frac{3}{2} , 0)$

• Triples, which are the solution of this task, are $(0,0,0) , ( \frac{1}{2} , - \frac{3}{2} , 0) , (1,-2,0) , (1,-1,-1)$ Trere are four of them.

By Vieta's, the condition is essentially $a+b+c=-a$ $ab+bc+ca=b$ $abc=-c.$ We rewrite the third equation as $c(ab+1)=0.$ We now have two cases to check.

Case 1: $c = 0$

In this case, the first equation gives $b=-2a$ and the second equation gives $ab=b\Rightarrow$ $b(a-1)=0.$ So either $b=0$ or $a=1$ . The first gives the solution $(0,0,0)$ and the second gives $(1,-2,0)$ .

Case 2: $ab=-1$

We substitute $b=-\dfrac{1}{a}$ into the equations to get $2a+c-\dfrac{1}{a}=0$ and $ca-1 +(1-c)\dfrac{1}{a}=0.$ Since $a\neq 0$ by the condition, we can multiply by $a$ to get the quadratics $2a^2+ca-1=0$ and $ca^2-a+(1-c)=0.$ These become $c=\dfrac{1-2a^2}{a}$ and $ca(a-1)=0.$ As stated earlier, we have $a\neq 0$ . Thus, the second equation gives either $c=0$ or $a=1$ .

The first gives the solution $(\dfrac{1}{\sqrt{2}}, -\sqrt{2}, 0)$ while the second gives the solution $(1,-1, -1)$ .

We can plug-in and see that all these solutions work, so that in total there are $\boxed{4}$ such triples of real numbers satisfying the conditions.