# Determine x + y

#### anemone

##### MHB POTW Director
Staff member
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

Staff member
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

##### MHB POTW Director
Staff member
Well done MarkFL for submitting a complete and correct solution in such a short period of time! I just love your approach so so much! #### Jester

##### Well-known member
MHB Math Helper
Here's my solution
Let $x = u +1$ and $y = v + 1$ giving

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

$u^3+v^3+2u+2v = 0$ 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

##### Well-known member
MHB Math Helper
Here's my solution
Let $x = u +1$ and $y = v + 1$ giving

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

$u^3+v^3+2u+2v = 0$ 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! -Dan

#### anemone

##### MHB POTW Director
Staff member
Here's my solution
Let $x = u +1$ and $y = v + 1$ giving

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

$u^3+v^3+2u+2v = 0$ 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!!! This is surely another impressive and great way to tackle this problem! 