# Thread: x^3 - y^3

1. ## x^3 - y^3

Suppose that x and y are two real numbers such that x-y=2 and x^2+y^2=8. Find x^3-y^3.

I remember I knew how to solve this at one point, but forgot over the summer.

2. Originally Posted by ceasar_19134
Suppose that x and y are two real numbers such that x-y=2 and x^2+y^2=8. Find x^3-y^3.

I remember I knew how to solve this at one point, but forgot over the summer.
x-y=2, now square both sides to get:

(x-y)^2=x^2+y^2-2xy=8-2xy=2^2

so xy=2.

Now multiply x-y and x^2+y^2 together to get:

(x-y)(x^2+y^2)=x^3-y^3+xy^2-yx^2=(x^3-y^3) + xy(y-x)

But the the term at the extreme left is equal to 2.8=16, and the term at the
extreme right is (x^3-y^3)-4, hence x^3-y^3=20.

RonL

Please check the algebra carefully to make sure it's correct.

3. Originally Posted by ceasar_19134
Suppose that x and y are two real numbers such that x-y=2 and x^2+y^2=8. Find x^3-y^3.

I remember I knew how to solve this at one point, but forgot over the summer.
Hi,

compare (x-y)^2 = x^2 - 2xy + y^2 = 4 and x^2 + y^2 = 8, thus 2xy = 4, thus xy = 2.

From x-y = 2 you get: x = y+2, therefore xy = 2 becomes: (y+2)y = 2

Simplify to: y^2 + 2y - 2 = 0

I leave the next steps to you. Substitute y in the equation x - y = 2 to calculate x.

I've got as final result to your problem x^3 - y^3 = 20.

tsch&#252;ss

EB

4. Hello, ceasar_19134!

A variation of the Captain's solution . . . and Earboth's.

Suppose that x and y are two real numbers such that:
. . [1] x - y .= .2 .and .[2] x² + y² .= .8

Find: x³ - y³

Square [1]: .(x - y)² .= . . . x² - 2xy + y² .= .4 . . (x² + y²) - 2xy .= .4

. . From [2] we have: .8 - 2xy .= .4 . . -2xy = -4 . . xy = 2 .[3]

Cube [1]: . (x - y)³ .= . . . x³ - 3x²y + 3xy² - y³ .= .8

. . We have: . x³ - y³ - 3xy(x - y) .= .8

From [2] and [3], we have: . x³ - y³ - 3(2)(2) .= .8

Therefore: . x³ - y³ .= .20

5. whats the next step after y^2+2y-2=o

6. Originally Posted by Rimas
whats the next step after y^2+2y-2=o
Hi, that's a quadratic equation. Use the formula, which you probably know, to solve this equation:

If the equation is: ax^2 + bx + c = 0 then the solution is: x = 1/(2a)(-b+- sqrt(b^2-4ac))

Use this formula: y= 1/(2*1)(-2+- sqrt(2^2-4*1*(-2)) = -1+- sqrt(3)

tschüss

EB