Algebra Problem, solving by rearranging

[PotentialE]

Homework Statement

Let x and y be real numbers with x+y=1 and (x² + y²)(x³ + y³) = 12. What is the value of x² + y² ?

Homework Equations

Sum of Cubes: (a³ + b³) = (a+b)(a²-ab+b²)

The Attempt at a Solution

I plugged in the sum of cubes to the equation that equals 12 to get:
(x² + y²)(x+y)(x²-xy+y²) = 12
and since (x+y) = 1,
(x² + y²)(x²-xy+y²) = 12
and therefore,
(x² + y²)(x²-xy+y²) = (x² + y²)(x³ + y³)
cancellation:
(x+y)(x²-xy+y²) = (x³ + y³) = 12

then I plugged (x³ + y³) for 12 to the original equation:
(x² + y²)(x³ + y³) = (x³ + y³)
and that means that (x² + y²) = 1, right?

 

[PotentialE]

The answer key that I just found said that the answer is 3...
Where did I go wrong?

 

[SammyS]

Homework Statement

Let x and y be real numbers with x+y=1 and (x² + y²)(x³ + y³) = 12. What is the value of x² + y² ?

Homework Equations

Sum of Cubes: (a³ + b³) = (a+b)(a²-ab+b²)

The Attempt at a Solution

I plugged in the sum of cubes to the equation that equals 12 to get:
(x² + y²)(x+y)(x²-xy+y²) = 12
and since (x+y) = 1,
(x² + y²)(x²-xy+y²) = 12
and therefore,
(x² + y²)(x²-xy+y²) = (x² + y²)(x³ + y³)

cancellation:
(x+y)(x²-xy+y²) = (x³ + y³) = 12

How did you arrive at the above line ?

then I plugged (x³ + y³) for 12 to the original equation:
(x² + y²)(x³ + y³) = (x³ + y³)
and that means that (x² + y²) = 1, right?

 

[PotentialE]

How did you arrive at the above line ?

A typo, my bad,
what I meant was:

and therefore,
(x2 + y2)(x+y)(x2-xy+y2) = (x2 + y2)(x3 + y3)
cancellations:
(x2 + y2)(x2-xy+y2) = (x2 + y2)(x3 + y3) = 12
(x2-xy+y2) = (x3 + y3)

I plugged in the sum of cubes to the equation that equals 12 to get:
(x2 + y2)(x+y)(x2-xy+y2) = 12
and since (x+y) = 1,
(x2 + y2)(x2-xy+y2) = 12
and therefore,
(x2 + y2)(x2-xy+y2) = (x2 + y2)(x3 + y3)
cancellation:
(x2-xy+y2) = (x3 + y3)

Now I realize that the last part is wrong, but where do I go from here?

 

[e^(i Pi)+1=0]

How did you get from

(x2 + y2)(x2-xy+y2) = 12
to
(x2 + y2)(x+y)(x2-xy+y2) = (x2 + y2)(x3 + y3) ?

I'm not saying it's wrong, I just don't follow it.

Anyway, after several pages of algebra, I also arrived at x2+y2=1
Are you sure you're looking at the right answer?

$x=1±\frac{1}{\sqrt{2}} , 1±\frac{\sqrt{5}i}{\sqrt{3}}$

edit: I plugged it into the original equation and it doesn't work, so I must have made a mistake somewhere. Odd that we both got the same answer..

 

[RoshanBBQ]

Homework Statement

Let x and y be real numbers with x+y=1 and (x² + y²)(x³ + y³) = 12. What is the value of x² + y² ?

Homework Equations

Sum of Cubes: (a³ + b³) = (a+b)(a²-ab+b²)

The Attempt at a Solution

I plugged in the sum of cubes to the equation that equals 12 to get:
(x² + y²)(x+y)(x²-xy+y²) = 12
and since (x+y) = 1,
(x² + y²)(x²-xy+y²) = 12
and therefore,
(x² + y²)(x²-xy+y²) = (x² + y²)(x³ + y³)
cancellation:
(x+y)(x²-xy+y²) = (x³ + y³) = 12

then I plugged (x³ + y³) for 12 to the original equation:
(x² + y²)(x³ + y³) = (x³ + y³)
and that means that (x² + y²) = 1, right?

Let
$$A = x^2 + y^2$$

$$(x^2 + y^2)(x^3 + y^3) = 12$$
$$A(x+y) (x^2-x y+y^2)=12$$
$$A(A-x y)=12$$
$$x+y = 1 \rightarrow (x+y)^2 = 1 \rightarrow x^2+y^2+2xy = 1 \rightarrow xy = \frac{1-x^2-y^2}{2}=\frac{1-A}{2}$$
$$A(A-\frac{1-A}{2})=12$$
$$A(\frac{3}{2}A-\frac{1}{2})=12$$
$$3A^2-A-24=0$$
You get two answers, but one is negative. We know the answer must be positive. The answer is 3.

 

[verty]

This is an olympiad type question, you need to play with it. Start with x = 1-y, then get formulas for (x^2 + y^2) and (x^3 + y^3). Look to make a substitution.

 

[PotentialE]

You get two answers, but one is negative. We know the answer must be positive. The answer is 3.

thanks for your help! that makes perfect sense