Answer

x+y=1 \implies (x+y)^2 = 1

xy = \dfrac{ (x+y)^2 -x^2-y^2}{2} = \dfrac{1-3}{2} = -1

x^3+y^3

= (x+y)(x^2-xy+y^2)

= 2-(-1 )=3

Steven Zheng posted 4 months ago

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