Question

Evaluate please (x 2 -y 2 ) 3 +(y 2 -z 2 ) 3 +(z 2 -x 2 ) 3 divided by (x-y) 3 +(y-z) 3 +(z-x) 3 Thanks in advance

Answer 1

For more, please comment and follow. Thanks!

Answer 2

Answer:

(x + y)(y + z)(z + x)

Step-by-step explanation:

We are supposed to Evaluate the value of the term,

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

We also know that from the property of sum of cubes,

$a^{3}+b^{3}+c^{3}=3abc\\if,\\a+b+c=0$

Therefore,

$(x^{2}-y^{2})^{3}+(y^{2}-z^{2})^{3}+(z^{2}-x^{2})^{3}=3(x^{2}-y^{2})(y^{2}-z^{2})(z^{2}-x^{2})\\because,\\(x^{2}-y^{2})+(y^{2}-z^{2})+(z^{2}-x^{2})=0$

Also,

In the same manner,

Using the same property of sum of cubes of the terms, we can say that,

$(x-y)^{3}+(y-z)^{3}+(z-x)^{3}=3(x-y)(y-z)(z-x)\\because,\\(x-y)+(y-z)+(z-x)=0$

Therefore, the term as evaluated is given as,

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

Therefore, the final evaluated term is given as,

(x + y)(y + z)(z + x)