Question

X3 (y-z)3 + y3 (z-x)3 + z3(x-y)3 factorise

Answer 1

Answer:

following answer

Step-by-step explanation:

Let a = x - y, b = y - z, c = z - x

Here, a + b + c = x - y + y - z + z - x = 0

Now if a + b + c = 0 then x3 + y3 + z3 = 3xyz

Hence,

(x - y)3 + (y - z)3 + (z - x)3 = 3(x - y) (y - z) (z - x).

HELPING HANDS✋:)

Answer 2

Given : $x^3(y-z)^3+y^3(z-x)^3+z^3(x-y)^3$

To find: Factorize

Step-by-step explanation:

Now

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

Now It can be written as

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

Now

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

As we know if $a+b+c=0$ then $a^3+b^3+c^3= 3abc$

Therefore

$a+b+c=x(y-z)+y(z-x)+z(x-y) \\\\\Rightarrow a+b+c=xy-xz+yz-xy+xz-yz=0$

Therefore

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

Hence, factorization is $3xyz(x-y)(y-z)(z-x)$