Question

factorise (x2-y2)3+(y2-z2)3(z2-x2)3/(x-y)3+(y-z) 3+(z-x) 3​

Answer 1

Answer:

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

Step-by-step explanation:

Formula used:

If a+b+c=0 then $a^3+b^3+c^3=3abc$

since$\:(x^2-y^2)+(y^2-z^2)+(z^2-x^2)=0$, using the above formula we get

$(x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3=3x^2y^2z^2$

similarly

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

Now,

$\frac{(x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3}{(x-y)^3+(y-z)^3+(z-x)^3}$

$=\frac{3x^2y^2z^2}{3xyz}$

$=xyz$