Solve the problem involving the cubic function

AI Thread Summary
The discussion revolves around proving that a specific value, denoted as b, is a root of a cubic function defined by the equation x^3 - 3bx^2 + 3cx - d. Participants emphasize the importance of demonstrating that f(b) equals zero to confirm b as a root, cautioning against assuming b is a root without proof. The roots are suggested to be in arithmetic progression, leading to the conclusion that they can be expressed as (b-r), b, and (b+r). The conversation highlights the need for careful algebraic manipulation to align the coefficients of the cubic polynomial with those of the given equation. Ultimately, the participants agree that the approach taken confirms b as a root under the specified conditions.
chwala
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Homework Statement
See attached
Relevant Equations
cubic equations and roots.
The problem and solution are posted... no. 8

I may need insight on common difference ...

1708681081283.png

1708681276104.png


In my lines i have,
Let the roots be ##(b), (b-1)## and ##(b+1)##.
Then,
##x^3-3bx^2+3cx-d = a(x-b(x-b+1)(x-b-1)##

##x^3-3bx^2+3cx-d= a(x^3-3bx^2+3b^2x-x-b^3+b)##
##a=1##.

Let

##f(x)=x^3-3bx^2+3b^2x-x-b^3+b##

Using Factor theorem,

##f(b)=b^3-3b^3+3b^3-b-b^3+b=0##

##f(b)=0## thus ##b## is a root of the cubic equation.

For the condition, we solve the equations,

##x^3-3bx^2+3cx-d= x^3-3bx^2+3b^2x-x-b^3+b##

##(3b^2-1)=3c##
##-b^3+b=-d##

##⇒b(b^2-(3b^2-3c))=d##

##b(-2b^2+3c)=d##

##2b^3-3bc+d=0##.

Any insight ...let me work on common difference later...
 
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You are asked to prove that b is a root; you cannot therefore assume that b is a root, or that the common difference is 1.

The cubic with leading cofficient 1 with p and p \pm q as roots is <br /> (x - p)((x - p)^2 - q^2) \equiv (x - p)^3 - q^2(x-p). To compare this to our cubic, complete the cube to obtain \begin{split}<br /> x^3 - 3bx^2 + 3cx - d &amp;= (x - b)^3 - 3(b^2 - c)x + b^3 - d \\<br /> &amp;= (x - b)^3 - 3(b^2 - c) \left(x - \frac{b^3 - d}{3(b^2 - c)}\right). \end{split} Now by comparison we must have \begin{split}<br /> p &amp;= b = \frac{b^3 - d}{3(b^2 - c)} \\<br /> q^2 &amp;= 3(b^2 - c).\end{split}
 
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Likes docnet, chwala and SammyS
pasmith said:
You are asked to prove that b is a root; you cannot therefore assume that b is a root, or that the common difference is 1.

The cubic with leading cofficient 1 with p and p \pm q as roots is <br /> (x - p)((x - p)^2 - q^2) \equiv (x - p)^3 - q^2(x-p). To compare this to our cubic, complete the cube to obtain \begin{split}<br /> x^3 - 3bx^2 + 3cx - d &amp;= (x - b)^3 - 3(b^2 - c)x + b^3 - d \\<br /> &amp;= (x - b)^3 - 3(b^2 - c) \left(x - \frac{b^3 - d}{3(b^2 - c)}\right). \end{split} Now by comparison we must have \begin{split}<br /> p &amp;= b = \frac{b^3 - d}{3(b^2 - c)} \\<br /> q^2 &amp;= 3(b^2 - c).\end{split}
I will go through your steps...cheers.
 
In order to show ##b## is a root, you need to show ##f(b)=0##. Notice that the only thing you can conclude from the assumption of arithmetic progression is that the roots are ##b-r,b, b+r##, or ##b, b+r, b+2r##. You're assuming ##r=1##.
 
WWGD said:
In order to show ##b## is a root, you need to show ##f(b)=0##. Notice that the only thing you can conclude from the assumption of arithmetic progression is that the roots are ##b-r,b, b+r##, or ##b, b+r, b+2r##. You're assuming ##r=1##.
yes, that was a mistake. I will re look into the steps...
 
WWGD said:
In order to show ##b## is a root, you need to show ##f(b)=0##. Notice that the only thing you can conclude from the assumption of arithmetic progression is that the roots are ##b-r,b, b+r##, or ##b, b+r, b+2r##. You're assuming ##r=1##.
While this did bear repeating, @pasmith covered this all very well in his earlier post.
 
SammyS said:
While this did bear repeating, @pasmith covered this all very well in his earlier post.
I think the emphasis on the ##r## part ... specifically let me state it as ##{d}## in terms of arithmetic progression was necessary...
 
pasmith said:
You are asked to prove that b is a root; you cannot therefore assume that b is a root, or that the common difference is 1.

The cubic with leading cofficient 1 with p and p \pm q as roots is <br /> (x - p)((x - p)^2 - q^2) \equiv (x - p)^3 - q^2(x-p). To compare this to our cubic, complete the cube to obtain \begin{split}<br /> x^3 - 3bx^2 + 3cx - d &amp;= (x - b)^3 - 3(b^2 - c)x + b^3 - d \\<br /> &amp;= (x - b)^3 - 3(b^2 - c) \left(x - \frac{b^3 - d}{3(b^2 - c)}\right). \end{split} Now by comparison we must have \begin{split}<br /> p &amp;= b = \frac{b^3 - d}{3(b^2 - c)} \\<br /> q^2 &amp;= 3(b^2 - c).\end{split}
Your steps are clear; though on the last line where you have,

##p=b=\dfrac{b^3-d}{3(b^2-c)}## is a bit confusing, this would imply that

##b=\dfrac{b^3-d}{3(b^2-c)}## which may not be true ... unless we have the condition that

##2b^3=3bc+d##.
 
WWGD said:
In order to show ##b## is a root, you need to show ##f(b)=0##. Notice that the only thing you can conclude from the assumption of arithmetic progression is that the roots are ##b-r,b, b+r##, or ##b, b+r, b+2r##. You're assuming ##r=1##.
Okay using what you and @pasmith hinted,
let the roots be ##(b-r), b## and ##(b+r)##
then we shall have ##(x-b)(x-b-r)9x-b+r)=(x-b)[(x-b)^2 -r^2]##

then,

##(x-b)^3-r^2(x-b)=x^3-3x^2b +3b^2x-r^2x-b^3+r^2b##

Let

##f(x)= (x-b)^3-r^2(x-b)=x^3-3x^2b +3b^2x-r^2x-b^3+r^2b##

then using Factor theorem,

##f(b)= b^3-3b^3 +3b^3-r^2b-b^3+r^2b=0##

##f(b)=0##

therefore ##b## is a root of the cubic equation.
 
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chwala said:
Okay using what you and @pasmith hinted,
let the roots be ##(b-r), b## and ##(b+r)##
then we shall have ##(x-b)(x-b-r)9x-b+r)=(x-b)[(x-b)^2 -r^2]##
Look!

If you start out saying that ##b## is one of the roots of the equation, then of course you will find that ##b## is a root of the equation.
 
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  • #11
WWGD said:
In order to show ##b## is a root, you need to show ##f(b)=0##. Notice that the only thing you can conclude from the assumption of arithmetic progression is that the roots are ##b-r,b, b+r##, or ##b, b+r, b+2r##. You're assuming ##r=1##.
Okay using what you and @pasmith hinted,
let the roots be ##(b-r), b## and ##(b+r)##
then we shall have ##(x-b)(x-b-r)9x-b+r)=(x-b)[(x-b)^2 -r^2]=(x-b)^3-r^2(x-b)=x^3-3x^2b +3b^2x-r^2x-b^3+r^2b##

Let ##f(x)= (x-b)^3-r^2(x-b)=x^3-3x^2b +3b^2x-r^2x-b^3+r^2b##

then,

##f(b)= b^3-3b^3 +3b^3-r^2b-b^3+r^2b=0##

##f(b)=0## using Factor theorem and therefore ##b## is a root of the cubic equation.
SammyS said:
Look!

If you start out saying that ##b## is one of the roots of the equation, then of course you will find that ##b## is a root of the equation.
Okay boss ... I'll replace (amend) the ##b## with another letter... will do this later. Cheers...
 
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  • #12
I've done some minor editing to the quoted parts:
chwala said:
Okay --- ... I'll replace (amend) the ##b## with another letter... will do this later. Cheers...

... using what you ( WWGD ) and @pasmith hinted,
let the roots be ##(b-r), b## and ##(b+r)##
then we shall have ##(x-b)(x-b-r)(x-b+r)=(x-b)[(x-b)^2 -r^2]=(x-b)^3-r^2(x-b)=x^3-3x^2b +3b^2x-r^2x-b^3+r^2b##

Let ##f(x)= (x-b)^3-r^2(x-b)##

##\quad\quad\quad\quad=x^3-3b\,x^2 +3b^2x-r^2x-b^3+r^2b##
You have done most of the necessary algebra. We might as well use the variable, ##p## rather than using ##b## in ensuring that we have an arithmetic progression. The variable, ##r##, is fine to use.

Then you have that ##(p-r),\, p## and ##(p+r)## are zeros of the following cubic polynomial.

##\displaystyle f(x)= (x-p)^3-r^2(x-p)##

##\displaystyle \quad\quad\quad=x^3-3p\,x^2 +3p^2x-r^2x-p^3+r^2p##

We need this polynomial to be identical to the given polynomial. Therefore, we must have that

##\displaystyle f(x)=x^3-3bx^2+cx-d ##

This implies that the corresponding coefficients must be equal respectively. In particular, the coefficient of ##x^2## in the first version of ##f(x)## must be equal to the coefficient of ##x^2## in the second version.

Edit: -to reword/replace the following sentence.
Does this show that ##b## is a root of the cubic equation in the given problem?

This does show that ##b## is a root of the given cubic equation, given that the roots are in arithmetic progression.
 
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  • #13
SammyS said:
I've done some minor editing to the quoted parts:

You have done most of the necessary algebra. We might as well use the variable, ##p## rather than using ##b## in ensuring that we have an arithmetic progression. The variable, ##r##, is fine to use.

Then you have that ##(p-r),\, p## and ##(p+r)## are zeros of the following cubic polynomial.

##\displaystyle f(x)= (x-p)^3-r^2(x-p)##

##\displaystyle \quad\quad\quad=x^3-3p\,x^2 +3p^2x-r^2x-p^3+r^2p##

We need this polynomial to be identical to the given polynomial. Therefore, we must have that

##\displaystyle f(x)=x^3-3bx^2+cx-d ##

This implies that the corresponding coefficients must be equal respectively. In particular, the coefficient of ##x^2## in the first version of ##f(x)## must be equal to the coefficient of ##x^2## in the second version.

Does this show that ##b## is a root of the cubic equation in the given problem?
@SammyS thanks so much for your invaluable contribution... got busy with work...will check on thread later boss.
 
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