If x and y are reals, and x+y=1 and (x^2+y^2)(x^3+y^3)=12, then what is the value of x^2 + y^2 ?
$\displaystyle x^3+y^3=(x+y)(x^2xy+y^2)=x^2xy+y^2$
$\displaystyle (x^2+y^2)(x^3+y^3)=(x^2+y^2)(x^2xy+y^2)=12$
Notice that if we square x+y=1 on both sides we get $\displaystyle x^2+y^2+2xy=1$ Now we have $\displaystyle 1+2xy=1$, $\displaystyle 2xy=0$. Therefore xy=0. This gives $\displaystyle (x^2+y^2)(x^2xy+y^2)=(x^2+y^2)^2=12$. So $\displaystyle (x^2+y^2)=\sqrt{12}$
Given: "x+y=1 and (x^2+y^2)(x^3+y^3)=12, then what is the value of x^2 + y^2 ? "
Solution: since (x + y) = 1, we can generate identities from it;
(x + y)^2 = x^2 + 2xy + y^2 = 1
x^2 + y^2 = 1  2xy    (1)
(x+y)^3 = x^3 + 3x^2 y + 3xy^2 + y^3 = 1
x^3 + y^3 +3xy(x + y) = 1
x^3 + y^3 + 3xy(1) = 1
x^3 + y^3 = 1  3xy    (2)
From the given expression: (x^2 + y^2)(x^3 + y^3) = 12, we equate (1) and (2)
we have (1  2xy)(1  3xy) = 12, expanding by FOIL Method
1  5xy + 6(xy)^2 = 12, rearranging
6(xy)^2  5xy  11 = 0, by quadratic formula,
xy = [5 +/ (25  4(6)9(11))^0.5]/2(6)
roots for xy: xy = (5 +/ 17)/12
(+) xy = 22/12 or
() xy = 12/12
for convenience SAY, xy = 1
then from (1)
x^2 + y^2 = 1  2xy = 1  2(1) = 3.

