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If \(x+y=4\) and \(x^2+y^2=8\), find \(x^3+y^3\).

 Jan 6, 2021
 #1
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(x+y)^2-x^2-y^2=xy=8

xy*(x+y)=32=x^2*y+x*y^2

(x+y)^3-3x^2*y-3y^2*x=x^3+y^3=-32

:D

 Jan 6, 2021
 #2
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We have that \(8=x^2+y^2=x^2+2xy+y^2-2xy=(x+y)^2-2xy=16-2xy\), therefore .\(xy=\frac{16-8}{2}=4\) Since \(x^3+y^3=(x+y)(x^2-xy+y^2)=(x+y)(x^2+y^2-xy)\), we can directly substitute in the numerical values for each algebraic expression. This gives us \(x^3+y^3=(4)(8-4)=\boxed{16}\).

 Jan 6, 2021

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