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if x+y=1 \[x^{3}+y^{3}=4 , find:x^{2}+y^{2}\]
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@mukushla
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!
ans. is 7
so i presume u didn't find a solution?
i got wrong ans.
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i got 3
\[x^3+y^3=(x+y)(x^2+y^2-xy)\]\[1=1.(x^2+y^2-xy)\]\[x^2+y^2-xy=1\]\[(x+y)^2-3xy=1\]\[(1)^2-3xy=1\]\[xy=0\]
sorry
if x+y=1 \[x^{3}+y^{3}=4 , find:x^{2}+y^{2}\]
\[x^3+y^3=(x+y)(x^2+y^2-xy)\]\[4=1.(x^2+y^2-xy)\]\[x^2+y^2-xy=4\]\[(x+y)^2-3xy=4\]\[xy=-1\]
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ok
\[ 1=(x+y)^2= x^2 + y^2 + 2 x y = x^2 +y^2 -2\\ x^2+y^2 =3 \]
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