Question

If $\frac{x}{2}=\frac{1}{\sqrt{7}+\sqrt{5}},\frac{y}{2}=\frac{1}{\sqrt{5}+\sqrt{3}},\frac{-z}{4}=\frac{1}{\sqrt{3}+\sqrt{7}},$ then the value of $x^3+y^3+z^3-2xyz$ is

Answer 1

$x+y+z=\frac{2\sqrt{15}+2\sqrt{35}+2\sqrt{9}+2\sqrt{21}+2\sqrt{49}+2\sqrt{15}+2\sqrt{35}+2\sqrt{21}-4\sqrt{25}-4\sqrt{15}-4\sqrt{35}-4\sqrt{21}}{(\sqrt{7}+\sqrt{5})(\sqrt{5}+\sqrt{3})(\sqrt{3}+\sqrt{7})}$

$x+y+z=\frac{6+14-20}{(\sqrt{7}+\sqrt{5})(\sqrt{5}+\sqrt{3})(\sqrt{3}+\sqrt{7})}=\frac{0}{(\sqrt{7}+\sqrt{5})(\sqrt{5}+\sqrt{3})(\sqrt{3}+\sqrt{7})}=0$

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

$x\ast y\ast z=\frac{(\sqrt{7}-\sqrt{5})(\sqrt{5}-\sqrt{3})(\sqrt{3}-\sqrt{7})\ast 2\ast 2\ast -4}{(\sqrt{7}+\sqrt{5})(\sqrt{5}+\sqrt{3})(\sqrt{3}+\sqrt{7})(\sqrt{7}-\sqrt{5})(\sqrt{5}-\sqrt{3})(\sqrt{3}-\sqrt{7})}$

$=\frac{(\sqrt{7}-\sqrt{5})(\sqrt{5}-\sqrt{3})(\sqrt{3}-\sqrt{7})\ast 2\ast 2\ast -4}{2\ast 2\ast -4}$

$x\ast y\ast z=(\sqrt{7}-\sqrt{5})(\sqrt{5}-\sqrt{3})(\sqrt{3}-\sqrt{7})$

Answer= $x\ast y\ast z=(\sqrt{7}-\sqrt{5})(\sqrt{5}-\sqrt{3})(\sqrt{3}-\sqrt{7})$