Question

If $x=\sqrt{2}-1,y=1-\sqrt{3}$ and $z=\sqrt{3}-\sqrt{2}$, find $x^3+y^3+z^3-xyz.$

Answer 1

Given $x=\sqrt{2}-1$, $y=1-\sqrt{3}$, $z=\sqrt{3}-\sqrt{2}$

here $x+y+z=\sqrt{2}-1+1-\sqrt{3}+\sqrt{3}-\sqrt{2}=0$

Hence $(x+y+z)=0$ ……….$\left(1\right)$

Now,

$(x+y+z{)}^3=x^3+y^3+z^3+3(x+y\left)\right(y+z\left)\right(z+x)$

from equation $\left(1\right)$ $z=-(x+y)$

then

$O=x^3+y^3+z^3+3(x+y)(y-x-y)(-x-y+x)$

$O=x^3+y^3+z^3+3(x+y)\left(xy\right)$

$O=x^3+y^3+z^3-3xyz$

$x^3+y^3+z^3-xyz=2xyz$

Hence $x^3+y^3+z^3-xyz=2(\sqrt{2}-1)(1-\sqrt{3})(\sqrt{3}-2)$

$=2(\sqrt{3}-2\sqrt{2}+1)$

$x^3+y^3+z^3-xyz=2\sqrt{3}-4\sqrt{2}+2$.