MHB Finding the Sum of Real Numbers Satisfying Cubic Equations

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The discussion focuses on finding the sum of real numbers x and y that satisfy the cubic equations x^3 - 3x^2 + 5x - 17 = 0 and y^3 - 3y^2 + 5y + 11 = 0. Participants share their solutions and approaches, with MarkFL receiving praise for a quick and accurate resolution. Other contributors express admiration for the problem-solving methods presented. The overall tone is supportive, highlighting the collaborative effort in tackling the mathematical challenge. The goal is to determine the value of x + y.
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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)
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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