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dtan5457
 one year ago
if x+y=1 and x^2+y^2=4, what is x^3+y^3
dtan5457
 one year ago
if x+y=1 and x^2+y^2=4, what is x^3+y^3

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dtan5457
 one year ago
Best ResponseYou've already chosen the best response.1weird precalc question i got today no idea how to do it

Empty
 one year ago
Best ResponseYou've already chosen the best response.5\[(x+y)^3 = x^3+3x^2y+3xy^2+y^3\] \[(x+y)^2 = x^2+2xy+y^2\] This should be all you need :)

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.1im gonna give it try.

Empty
 one year ago
Best ResponseYou've already chosen the best response.5Sounds good, I'll give you some more hints if you need them later. :D

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.1the formula is just the most simplfied versions of x^3+y^3=(x+y)(x^2xy+y^2) right?

Hero
 one year ago
Best ResponseYou've already chosen the best response.1you have the formula. All you have to do is just input the given Info

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.1im just confused on whether im suppose to use the x^3+Y^2 formula or (x+y)^3

Hero
 one year ago
Best ResponseYou've already chosen the best response.1(x+y)(x^2xy+y^2)=(1)(4xy)

Empty
 one year ago
Best ResponseYou've already chosen the best response.5\[(x+y)^3 = x^3+3x^2y+3xy^2+y^3\] \[(x+y)^3 = x^3+y^3 + 3xy(x+y)\] \[(x+y)^3  3xy(x+y)= x^3+y^3\] So I arranged it so we have \(x^3+y^3\) which is what we want, and we know that \(x+y=1\) so we plug it in now: \[1^3  3xy(1)=13xy = x^3+y^3\] What's xy? Let's use this: \[(x+y)^2 = x^2+2xy+y^2\] \[\frac{(x+y)^2  (x^2+y^2)}{2} = xy\] \[\frac{1^2  4}{2} =\frac{3}{2} = xy\] Now we can plug it in: \[13 \frac{3}{2} = x^3+y^3\]

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.1im seeing what you did but if I had started with the formula (x+y)(x^2xy+y^2) if i plug in from there I get 4xy? could i continue form here?

Empty
 one year ago
Best ResponseYou've already chosen the best response.5Yeah It looks like that'd probably work too

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.1im not sure what to do from there though

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.1do i use the same formula ?

Hero
 one year ago
Best ResponseYou've already chosen the best response.1Use(x+y)^2=x^2+2xy+y^2 from there

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.1Oh I think I got it. 1^2=4+2xy 14=3 3/2=2xy xy=3/2 1(43/2)=x^3+y^3

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.1can you check for me? and how would you know which formula to use? it seems like you would need more than 1 to compute all the variables

Hero
 one year ago
Best ResponseYou've already chosen the best response.1you will end up with ×y =3/2 so (1) (4xy)=4(3/2)= 4+3/2= x^3 + y^3

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.1oh, dumb negative mistake by me. this is kind of confusing because i never substituted formulas to solve questions like this. normally, how would you know which formulas to use?

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.1like how we used x^3+y^3 then to (x+y)^2?

Hero
 one year ago
Best ResponseYou've already chosen the best response.1Try to reason through it using the Known formulas

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.1you are right. this is only my first question with this, i just need to practice more. thank you both so much.
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