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If x-y=1 and x^2+y^2=7, find x^3-y^3.

 

Thanks in advance.  

 May 12, 2020
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\((x-y)=1\) square both sides

\(x^2-2xy+y^2=1\) (1)

given \(x^2+y^2=7\) substitute into (1)

\(7-2xy=1\)

Solve for \(xy\)

\(-2xy=-6\)

\(xy=3\)

\(x^3-y^3=(x-y)(x^2+xy+y^2)\) (The difference of two cubes) 

Substitute all givens and the value of \(xy\)

\((1)(7+3)\)

\(1*10=10\)

Thus \(x^3-y^3=10\)

 May 12, 2020

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