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x^3 + y^3

Date: 10/17/2002 at 17:46:31
From: Tjoseph8
Subject: x^3 + y^3

Given:  (1)   x + y = 1
        (2)   x^2 + y^2 = 4

Question:    x^3 + y^3 = ?

To solve this problem, can you solve (2) for x: x^2 = 4 - y^2
                           and       (1) for x: x = 1 - y

then set  4 - y^2 = 1 - y  and factor?

Date: 10/17/2002 at 19:08:23
From: Doctor Greenie
Subject: Re: x^3 + y^3

Hello -

You can go that route; but the resulting quadratic expression is not 
factorable and so you have to use the quadratic formula. If you are 
careful with the algebra, however, you can arrive at the answer to 
your problem by that method.

In the last equation you are equating two expressions for x^2, so the 
right-hand side should be (1 - y)^2 instead of just 1 - y. So your 
last equation will read

  4-y^2 = (1-y)^2
  4-y^2 = 1-2y+y^2
  2y^2-2y-3 = 0

The quadratic formula then gives

       2 + or - sqrt(28)     1 + or - sqrt(7)
  y = ------------------- = ------------------
               4                     2

To finish the problem by this method, you next use the fact that 
x+y = 1 to find the value for x; then you find the cube of x and the 
cube of y and add them together to evaluate x^3+y^3.

Using your method, you actually find values for x and y. The problem 
doesn't require that you do so; and solving it by that method requires 
more complicated algebra than is necessary.

The problem only asks for the value of x^3 + y^3. There is another 
general method for attacking problems like this.

The general idea of this alternate method is to multiply one or more 
of the given expressions together to get an expression containing the 
terms x^3 and y^3 and see if you can evaluate the value of x^3 + y^3 
from the resulting equation(s). Unless you have a lot of experience 
solving problems of this type, it is unlikely that you will know 
whether a particular path you start down using this method 
will lead you anywhere useful - you just need to go ahead and try 
those paths and see what happens.

In this example, we have two given equations:

(1) x+y = 1
(2) x^2+y^2 = 4

One way we can get an expression containing the desired terms is to 
multiply the expressions in (1) and (2) together:

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

Substituting the known values for (x+y) and (x^2+y^2), we have

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

And we have obtained the following equation for x^3+y^3:

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

Unfortunately, we don't know the value of xy, so this by itself hasn't 
solved the problem for us.

So let's go back and find another way of obtaining an expression 
containing the desired terms. One other way to do this is to multiply 
(1) by itself three times:

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

Again substituting the known values for (x+y) and (x^2+y^2), we have

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

And we have obtained another equation for x^3+y^3:

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

Again, we don't know the value of xy, so this too, by itself, hasn't 
solved the problem for us.

But there is a path to the solution to be found among our results.  
From equations (3) and (4), we have

  4-xy = 1-3xy
  2xy = -3
  xy = -3/2

Now we DO know the value of xy, so we can find the value of x^3 + y^3 
using that value for xy in either equation (3) or (4):

  x^3+y^3 = 4-xy = 4-(-3/2) = 4+3/2 = 11/2

This is the answer you will get if you continue down the path you 
started on originally.

I hope all this helps.  Please write back if you have any further 
questions about any of this.

- Doctor Greenie, The Math Forum 
Associated Topics:
High School Basic Algebra

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