1. ## Having Trouble

Hi all,

I am just new to this site and it looks like such an awesome resource.

Can anyone help me with this???

If x + y = 1 and x³ + y³ = 19, find the value of x² + y²

2. Originally Posted by Joel
Hi all,

I am just new to this site and it looks like such an awesome resource.

Can anyone help me with this???

If x + y = 1 and x³ + y³ = 19, find the value of x² + y²
$\displaystyle \left\{\begin{array}{lclcr} x&+&y&=&1\\ x^3&+&y^3&=&19 \end{array}\right.$

$\displaystyle \Rightarrow x^3+(1-x)^3=19,$

and after expanding and simplifying,

$\displaystyle 3x^2-3x-18=0$

$\displaystyle \Rightarrow3(x-3)(x+2)=0.$

You should be able to finish.

3. Hello, Joel!

Welcome aboard!

I have a back-door approach . . .

If .$\displaystyle \begin{array}{cccc}x + y &=& 1 & {\color{blue}[1]} \\ x^3 + y^3 &=& 19 & {\color{blue}[2]} \end{array}$ . find the value of $\displaystyle x^2+y^2.$

Cube [1]: .$\displaystyle (x+y)^3 \:=\:1^3\quad\Rightarrow\quad x^3 + 3x^2y + 3xy^2 + y^3 \:=\:1$

$\displaystyle \text{We have: }\;\underbrace{(x^3 + y^3)}_{\text{This is 19}} + \;3xy\underbrace{(x+y)}_{\text{This is 1}} \:=\:1 \quad\Rightarrow\quad 19 \;+ \;3xy \:=\:1 \quad\Rightarrow\quad xy \:=\:-6$

Square [1]: .$\displaystyle (x+y)^2 \:=\:1^2 \quad\Rightarrow\quad x^2+2xy + y^2 \:=\:1$

$\displaystyle \text{We have: }\;x^2 + y^2 + 2(xy) \:=\:1$
. . . . . . . . . . . . . . $\displaystyle \uparrow$
. . . . . . . . . . . . $\displaystyle ^{\text{This is }\text{-}6}$

Therefore: .$\displaystyle x^2+y^2 - 12 \:=\:1 \quad\Rightarrow\quad \boxed{x^2+y^2 \:=\:13}$

This method is guarenteed to impress/surprise/terrify your teacher.
.

4. Originally Posted by Soroban
Hello, Joel!

Welcome aboard!

I have a back-door approach . . .
I like your solution. I wish I had thought about the problem a bit more before going down the obvious path.