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 http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum