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Mathematics 21 Online
OpenStudy (anonymous):

if x+y=1 \[x^{3}+y^{3}=4 , find:x^{2}+y^{2}\]

OpenStudy (anonymous):

@mukushla

OpenStudy (anonymous):

This question does not really have an answer, cause the answer will always remain in terms of x and y. I will not give you an answer and let you do it yourself x^3 + y^3 = (x+y)(x^2 + y^2 -xy) now put the value of x+y and substitute xy it in x^2 + y^2 = 1-2xy. Let me know if you don't get the answer!

OpenStudy (anonymous):

ans. is 7

OpenStudy (anonymous):

so i presume u didn't find a solution?

OpenStudy (anonymous):

i got wrong ans.

OpenStudy (anonymous):

i got 3

OpenStudy (anonymous):

\[x^3+y^3=(x+y)(x^2+y^2-xy)\]\[1=1.(x^2+y^2-xy)\]\[x^2+y^2-xy=1\]\[(x+y)^2-3xy=1\]\[(1)^2-3xy=1\]\[xy=0\]

OpenStudy (anonymous):

sorry

OpenStudy (anonymous):

if x+y=1 \[x^{3}+y^{3}=4 , find:x^{2}+y^{2}\]

OpenStudy (anonymous):

\[x^3+y^3=(x+y)(x^2+y^2-xy)\]\[4=1.(x^2+y^2-xy)\]\[x^2+y^2-xy=4\]\[(x+y)^2-3xy=4\]\[xy=-1\]

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

\[ 1=(x+y)^2= x^2 + y^2 + 2 x y = x^2 +y^2 -2\\ x^2+y^2 =3 \]

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