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If x+y=9 and xy=18, what is the value of x^3+y^3?

 Jun 1, 2022

Best Answer 

 #2
avatar+2448 
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We have \(x^3 + y^3\). Dividing this by \((x+y)\), we get \((x^2 + y^2 - xy)\),

 

Now, we recall the identity: \(x^2 + y^2 = (x+y)^2 - 2xy\)

 

Substituting this in gives us: \((x+y)((x+y)^2 - 3xy)\)

 

Now, plugging in the values we know, we have: \(9 \times (81 - 3 \times 18) = \color{brown}\boxed{243}\)

 Jun 2, 2022
 #1
avatar+14084 
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What is the value of x^3+y^3?

 

Hello Guest!

 

\(x+y=9\\ xy=18\\ y=\frac{18}{x}=9-x\\ 18=9x-x^2\\ x^2-9x+18=0\\ x=4.5\pm \sqrt{4.5^2-18}=4.5\pm 1.5\)

\(x\in \{6,\ 3\} \\ y\in \{3,\ 6\}\\ \color{blue}x^3+y^3= 216+27=243\)

laugh  !

 Jun 1, 2022
 #2
avatar+2448 
0
Best Answer

We have \(x^3 + y^3\). Dividing this by \((x+y)\), we get \((x^2 + y^2 - xy)\),

 

Now, we recall the identity: \(x^2 + y^2 = (x+y)^2 - 2xy\)

 

Substituting this in gives us: \((x+y)((x+y)^2 - 3xy)\)

 

Now, plugging in the values we know, we have: \(9 \times (81 - 3 \times 18) = \color{brown}\boxed{243}\)

BuilderBoi Jun 2, 2022

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