Evaluate the value of the product

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Discussion Overview

The discussion revolves around evaluating a product involving pairs of roots from a system of equations defined by cubic expressions. The focus includes the verification of the equations and the implications of potential typographical errors in the problem statement.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Homework-related

Main Points Raised

  • Post 1 presents the original equations and the task of evaluating a specific product of differences between the roots.
  • Post 2 suggests a correction to the first equation, proposing it should be $a^3-3a^2b=2005$ instead.
  • Post 3 acknowledges a possible typo in the equations, indicating that another participant had previously pointed this out, and expresses regret for the confusion.
  • Post 4 claims to have solved the problem but does not provide details in the post.
  • Post 5 reiterates the solution claim and emphasizes the forum's guideline against linking to external sites for solutions.

Areas of Agreement / Disagreement

Participants do not reach consensus on the correctness of the original equations, with some suggesting a typo while others proceed with the original formulation. The discussion remains unresolved regarding the implications of these potential errors on the evaluation task.

Contextual Notes

The discussion highlights the importance of precise definitions in mathematical problems, as well as the potential for miscommunication regarding problem statements. The resolution of the equations and the validity of the proposed solutions remain contingent on clarifying these issues.

anemone
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The three pairs of roots $(a,\,b)$ that satisfy $a^3-3ab^2=2005$ and $b^3-3b^2a=2004$ are $(a_1,\,b_1),\,(a_2,\,b_2),\,(a_3,\,b_3)$.

Evaluate $\left(\dfrac{b_3-a_3}{b_3}\right)\left(\dfrac{b_2-a_2}{b_2}\right)\left(\dfrac{b_1-a_1}{b_1}\right)$.
 
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$a^3-3ab^2=2005---(1)$
$b^3-3b^2a=2004---(2)$
(2)-(1) and simplify we get :$(b-a)^3=-1,\,\ or\\, (b-a)=-1--(3)$
put (3) to (2) we get :$2b^3+3b^2+2004=0---(4)$
$b_3-a_3=b_2-a_2=b_1-a_1=b-a=-1$
and $b_1b_2b_3=-1002$
$\therefore $ the answer =$\dfrac {1}{1002}$
 
I have a question to Albert, because I do not quite understand his deduction:
The equations
\[a^3-3ab^2 = 2005\: \: \: \: \: (1). \\\\ b^3-3b^2a = 2004\: \: \: \: \: (2).\]
lead to:
\[b^3-a^3 = -1 \: \: \:\: (3).\]
and not to:
\[(b-a)^3 = -1\]
- because there is not the term $3a^2b$ in either (1) or (2)??
Thanks for clearing the matter
 
I think (1) should be :$a^3-3a^2b=2005$
 
Ops...I previously received a PM that notified me of the possible typo that I could have made, but I thought he mentioned of the constant value, apparently Albert was right, the first equation has a typo in it, and his intuition was right as well. :o

Sorry for both the late reply and late clarification post. :(
 
kaliprasad said:

In our guidelines for posting solutions, we state:

"Please do not give a link to another site as a means of providing a solution, either by the author of the topic posted here, or by someone responding with a solution."

We don't want our readers to have to follow a link to view a solution. :D
 

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