Finding the Sum of Real Numbers Satisfying Cubic Equations

[anemone]

The real numbers \(\displaystyle x\) and \(\displaystyle y\) satisfy \(\displaystyle x^3-3x^2+5x-17=0\) and \(\displaystyle y^3-3y^2+5y+11=0\). Determine the value of \(\displaystyle x+y\).

 

[MarkFL]

Spoiler

Let \(\displaystyle z=x+y\) and so the first equation becomes:

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

Which becomes:

\(\displaystyle z^3-3yz^2+3y^2z-y^3-3z^2+6yz-3y^2-5y+5z-17=0\)

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

Using the second equation, this becomes:

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

Now, the second equation may be written:

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

Which becomes:

\(\displaystyle z^3-3xz^2+3x^2z-x^3-3z^2+6xz-3x^2-5x+5z+11=0\)

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

Using the first equation, this becomes:

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

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

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

Hence:

\(\displaystyle z=x+y=2\)

 

[anemone]

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)

 

[Mathwebster]

Here's my solution

Spoiler

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 $x + y = u+v+2 = 2$.

 

[topsquark]

Here's my solution

Spoiler

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

 

[anemone]

Here's my solution

Spoiler

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)