Please help, stuck on seemingly simple problem

[Jt00]

If x+y=1 and xy=2, then find x^3+y^3.
I've gotten it narrowed down to: (x+y)(x^2-xy+y^2)
which simplifies to: 1(x^2-2+y^2)
and then to: (x^2)-2+(y^2) (parentheses for clarity)
What else can be done?

 

[Goku]

(x+y)^2 = x^2 + 2xy + y^2
1^2 - 2xy = x^2 + y^2

Now work from there to get a value for x^2 + y^2...

 

[soroban]

Hello, Jt00!

There's an even faster way . . .

\(\displaystyle \text{If }x+y=1\text{ and }xy=2\text{, then find }x^3+y^3.\)

We have: .\(\displaystyle x + y \:=\:1\)

Cube: . \(\displaystyle (x+y)^3 \:=\:1^3\)

We have:. . \(\displaystyle x^3 + 3x^2y + 3xy^2 + y^3 \:=\:1\)

. . . . . . . . \(\displaystyle x^3 + y^3 + 3\underbrace{(xy)}_{\downarrow}\underbrace{(x+y)}_{\swarrow\quad\;} \:=\:1\)
. . . . . . . . .\(\displaystyle x^3 + y^3 + 3(2)(1) \:=\:1\)

. . . . . . . . . . \(\displaystyle x^3 + y^3 + 6 \:=\:1\)

. . . . . . . . . . . \(\displaystyle x^3 + y^3 \:=\:\text{-}5\)

 

[Jt00]

Thanks! I knew I was forgetting something... btw, how do you write in the font where you can have sub and superscripts and math symbols?