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.

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- Oct 2nd 2006, 07:18 AM #1

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- Oct 2nd 2006, 07:30 AM #2

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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.

- Oct 2nd 2006, 07:47 AM #3
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.

tschüss

EB

- Oct 2nd 2006, 08:30 AM #4

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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 .= .2² . → . 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 .= .2³ . → . 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**

- Oct 2nd 2006, 08:32 AM #5

- Oct 2nd 2006, 08:43 AM #6

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