cant seem to figure out how to do this problem

Suppose x+y= 1 and x^2+y^2=4

find with algebraic explaination the value of x^3+y^3

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- Apr 19th 2007, 04:09 PM #1

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- Apr 19th 2007, 04:11 PM #2
we can use the sum of two cubes formula:

Note that x^3 + y^3 = (x + y)(x^2 - xy + y^2) = (x + y)(x^2 + y^2 - xy)

now we can plug in x+y= 1 and x^2+y^2=4

so x^3 + y^3 = (1)(4 - xy)

=> x^3 + y^3 = 4 - xy

if we wanted to, we could solve for x and y directly, but the question didn't ask for that

EDIT: Oh sorry, we're not done yet. We need to find the value of x^3 + y^3, so we don't want xy in the expression. let's find a numerical value for xy.

we are told x + y = 1

=> (x + y)^2 = 1^2

=> x^2 + 2xy + y^2 = 1 .........but x^2 + y^2 = 4

=> 4 + 2xy = 1

=> 2xy = -3

=> xy = -3/2

so x^3 + y^3 = 4 - xy = 4 + 3/2 = 11/2 or 5 1/2

- Apr 19th 2007, 06:32 PM #3

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Hello, mcdaking84!

Another approach . . .

Given: . .x + y .= .1 -**[1]**

. .and: .x² + y² .= .4 .**[2]**

Find the value of: .x³ + y³

Square [1]: .(x + y)² .= .1² . → . x² + 2xy + y² .= .1 . → . 2xy + (x² + y²) .= .1

From [2], we have: .2xy + 4 .= .1 . → . xy = -3/2 .**[3]**

Cube [1]: .(x + y)³ .= .1³ . → . x³ + 3x²y + 3xy² + y³ .= .1

Then we have: . . . x³ + y³ + 3xy(x + y) .= .1

. . . . . . . . . . . . . . . . . . . . . . ↓ . . .↓

From [3] and [1]: .x³ + y³ + 3(-3/2)(1) .= .1

Therefore: .x³ + y³ - 9/2 .= .1 . → . x³ + y³ .= .11/2