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If $x-y=6$ and $x^2+y^2=24$, find $x^3-y^3$.

 May 13, 2018
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Hi Guest!

 

We use: \(x^3-y^3=(x-y)(x^2+xy+y^2) \)

 

We already know \( (x-y)\) and \(x^2+y^2\)

 

We can plug the known values in, and we get:

 

\(6(24+xy)=x^3-y^3\)

 

To solve for xy, we use: \((x-y)^2=x^2-2xy+y^2\)

 

This means that \(6^2=24-2xy\)

 

Solving for xy, we get:

 

\(36=24-2xy\\ 12=-2xy\\ xy=-6\)

 

Now we can plug this into the original expression:
\(6(24-6)=x^3-y^3\\ x^3-y^3=108\)

 

I hope this helped,

 

Gavin

 May 13, 2018

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