Question

factorise ( x-3y)^3+(3y-7z)^3+(7z-x)^3​

Answer 1

Let a = x-3y ; b = 3y-7z ; c = 7z-x

now,

a + b + c=x-3y + 3y-7z + 7z-x = 0

Since, a + b + c = 0,

then we have,

a^3+b^3+c^3=3abc

HENCE,

(x-3y)^3+(3y-7z)^3+(7z-x)^3=3(x-3y) (3y-7z) (7z-x)

Hope this helps☺