Finding the Sum of Real Numbers Satisfying Cubic Equations

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Discussion Overview

The discussion revolves around finding the sum of real numbers \(x\) and \(y\) that satisfy two cubic equations: \(x^3-3x^2+5x-17=0\) and \(y^3-3y^2+5y+11=0\). The focus is on various proposed solutions to this problem.

Discussion Character

  • Homework-related, Mathematical reasoning

Main Points Raised

  • One participant presents a solution to the equations.
  • Another participant expresses admiration for the first solution, indicating it was complete and correct.
  • Multiple participants share their own solutions, suggesting different approaches to the problem.
  • Expressions of appreciation for the contributions of others are noted, indicating a collaborative atmosphere.

Areas of Agreement / Disagreement

There is no clear consensus on a single solution, as multiple participants have proposed their own methods without resolving which is the most effective or correct.

Contextual Notes

Details on the specific methods used in the proposed solutions are not provided, leaving the discussion open to interpretation and further exploration.

Who May Find This Useful

Participants interested in solving cubic equations or those looking for collaborative problem-solving approaches in mathematics may find this discussion beneficial.

anemone
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The real numbers $$x$$ and $$y$$ satisfy $$x^3-3x^2+5x-17=0$$ and $$y^3-3y^2+5y+11=0$$. Determine the value of $$x+y$$.
 
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Let $$z=x+y$$ and so the first equation becomes:

$$(z-y)^3-3(z-y)^2+5(z-y)-17=0$$

Which becomes:

$$z^3-3yz^2+3y^2z-y^3-3z^2+6yz-3y^2-5y+5z-17=0$$

$$z^3-3yz^2+3y^2z-3z^2+6yz-3y^2+5z-6-\left(y^3+5y+11 \right)=0$$

Using the second equation, this becomes:

(1) $$z^3-3yz^2+3y^2z-3z^2+6yz-6y^2+5z-6=0$$

Now, the second equation may be written:

$$(z-x)^3-3(z-x)^2+5(z-x)+11=0$$

Which becomes:

$$z^3-3xz^2+3x^2z-x^3-3z^2+6xz-3x^2-5x+5z+11=0$$

$$z^3-3xz^2+3x^2z-3z^2+6xz-3x^2+5z-6-\left(x^3+5x-17 \right)=0$$

Using the first equation, this becomes:

(2) $$z^3-3xz^2+3x^2z-3z^2+6xz-6x^2+5z-6=0$$

Adding (1) and (2), and simplifying, we have:

$$(z-2)\left(3\left(x^2+y^2 \right)-\left(z^2+2z-6 \right) \right)=0$$

Hence:

$$z=x+y=2$$
 
Well done MarkFL for submitting a complete and correct solution in such a short period of time!(Clapping)

I just love your approach so so much!(Inlove)
 
Here's my solution
Let $x = u +1$ and $y = v + 1$ giving

$u^3+2u-14=0$ and $v^3+2v+14=0$.

Adding these two gives

$u^3+v^3+2u+2v = 0$ or $ (u+v)(u^2-uv+v^2+2) = 0$.

Since the second term is strictly positive then $u+v = 0$ so that $x + y = u+v+2 = 2$.
 
Jester said:
Here's my solution
Let $x = u +1$ and $y = v + 1$ giving

$u^3+2u-14=0$ and $v^3+2v+14=0$.

Adding these two gives

$u^3+v^3+2u+2v = 0$ or $ (u+v)(u^2-uv+v^2+2) = 0$.

Since the second term is strictly positive then $u+v = 0$ so that $x + y = u+v+2 = 2$.
(Shakes his head.) Wow! (Bow)

-Dan
 
Jester said:
Here's my solution
Let $x = u +1$ and $y = v + 1$ giving

$u^3+2u-14=0$ and $v^3+2v+14=0$.

Adding these two gives

$u^3+v^3+2u+2v = 0$ or $ (u+v)(u^2-uv+v^2+2) = 0$.

Since the second term is strictly positive then $u+v = 0$ so that $x + y = u+v+2 = 2$.

Thanks for participating, Jester and WOW!(Clapping) This is surely another impressive and great way to tackle this problem!(Nerd)
 

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