Dizzy algebra

Let $(x-y)=a\quad (y-z)=b\quad (z-x)=c$

the question is of the form $\frac { { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 } }{ abc }$

$(x-y)+(y-z)+(z-x)=0$

so $a+b+c=0$

$(a+b)=-c$

${ (a+b) }^{ 3 }={ (-c) }^{ 3 }$

${ a }^{ 3 }+{ b }^{ 3 }+3ab(a+b)={ -c }^{ 3 }$

because $(a+b)=-c$ we get

${ a }^{ 3 }+{ b }^{ 3 }+3ab(-c)={ -c }^{ 3 }$

${ a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }=3abc$

$\frac { { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 } }{ abc } =3$

So the answer is $3$

just put x=1,y=2 and z=3 and solve you get the answer

Shortcut :- Just plug in the random values for the variables (x,y,z) => (1,2,3) => {(1-2)^3+(2-3)^3+(3-1)^3}/(1-2)(2-3)(3-1) => (-1-1+8)/(-1)(-1)(2) => 3

Let $(x-y)=a \quad (y-z)=b \quad (z-x)=c$ .

Then, $\frac{(x-y)^{3}+(y-z)^{3}+(z-x)^{3}}{(x-y)(y-z)(z-x)}=\frac{a^{3}+b^{3}+c^{3}}{abc}$

$\frac{a^{3}+b^{3}+c^{3}}{abc}=\frac{a^{3}+b^{3}+c^{3}-3abc}{abc}+3$

$\frac{a^{3}+b^{3}+c^{3}}{abc}=\frac{(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ac)}{abc}+3$

Because $\quad a+b+c=(x-y)+(y-z)+(z-x)=0$

Thus, the value of $\frac{(x-y)^{3}+(y-z)^{3}+(z-x)^{3}}{(x-y)(y-z)(z-x)}$ is $\boxed{3}$

putting a = y-z, b = z-x, c = x-y we get a+b+c=0 we get equation is (a3+b3+c3)/abc = (a3+b3+c3-3abc+3abc)/abc =((a+b+c)(a2+b2+c2-ab-bc-ca) +3abc)/abc = 3abc/abc=3