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)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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