+2666 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: \(9 \times (81 - 3 \times 18) = \color{brown}\boxed{243}\)
+14913 What is the value of x^3+y^3?
Hello Guest!
\(x+y=9\\ xy=18\\ y=\frac{18}{x}=9-x\\ 18=9x-x^2\\ x^2-9x+18=0\\ x=4.5\pm \sqrt{4.5^2-18}=4.5\pm 1.5\)
\(x\in \{6,\ 3\} \\ y\in \{3,\ 6\}\\ \color{blue}x^3+y^3= 216+27=243\)
!
+2666 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: \(9 \times (81 - 3 \times 18) = \color{brown}\boxed{243}\)
