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dtan5457

  • one year ago

if x+y=1 and x^2+y^2=4, what is x^3+y^3

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  1. dtan5457
    • one year ago
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    weird precalc question i got today no idea how to do it

  2. dtan5457
    • one year ago
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    @Nnesha @iambatman

  3. dtan5457
    • one year ago
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    @StudyGurl14

  4. Empty
    • one year ago
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    \[(x+y)^3 = x^3+3x^2y+3xy^2+y^3\] \[(x+y)^2 = x^2+2xy+y^2\] This should be all you need :)

  5. dtan5457
    • one year ago
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    im gonna give it try.

  6. Empty
    • one year ago
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    Sounds good, I'll give you some more hints if you need them later. :D

  7. dtan5457
    • one year ago
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    the formula is just the most simplfied versions of x^3+y^3=(x+y)(x^2-xy+y^2) right?

  8. Hero
    • one year ago
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    you have the formula. All you have to do is just input the given Info

  9. dtan5457
    • one year ago
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    im just confused on whether im suppose to use the x^3+Y^2 formula or (x+y)^3

  10. Hero
    • one year ago
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    (x+y)(x^2-xy+y^2)=(1)(4-xy)

  11. Empty
    • one year ago
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    \[(x+y)^3 = x^3+3x^2y+3xy^2+y^3\] \[(x+y)^3 = x^3+y^3 + 3xy(x+y)\] \[(x+y)^3 - 3xy(x+y)= x^3+y^3\] 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: \[1^3 - 3xy(1)=1-3xy = x^3+y^3\] 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: \[1-3 \frac{-3}{2} = x^3+y^3\]

  12. dtan5457
    • one year ago
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    im seeing what you did but if I had started with the formula (x+y)(x^2-xy+y^2) if i plug in from there I get 4-xy? could i continue form here?

  13. Empty
    • one year ago
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    Yeah It looks like that'd probably work too

  14. dtan5457
    • one year ago
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    im not sure what to do from there though

  15. dtan5457
    • one year ago
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    do i use the same formula ?

  16. Hero
    • one year ago
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    Use(x+y)^2=x^2+2xy+y^2 from there

  17. Hero
    • one year ago
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    And solve for xy

  18. dtan5457
    • one year ago
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    Oh I think I got it. 1^2=4+2xy 1-4=-3 -3/2=2xy xy=-3/2 1(4-3/2)=x^3+y^3

  19. dtan5457
    • one year ago
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    can 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

  20. Hero
    • one year ago
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    you will end up with ×y =-3/2 so (1) (4-xy)=4-(-3/2)= 4+3/2= x^3 + y^3

  21. dtan5457
    • one year ago
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    oh, 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?

  22. dtan5457
    • one year ago
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    like how we used x^3+y^3 then to (x+y)^2?

  23. Hero
    • one year ago
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    Try to reason through it using the Known formulas

  24. dtan5457
    • one year ago
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    you 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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