Question

factorize
(p-q)^3+(q-r)^3+(r-p)^3
solve fast

Answer 1

if a+b+c = 0
then a³+b³+c³ = 3abc
here , a = p-q
b= q-r , c= r-p
a+b+c => (p-q)+(q-r)+(r-p)
p-q+q-r+r-p = 0
so , (p-q)³+(q-r)³+(r-p)³ = 3(p-q)(q-r)(r-p)
hope this helps

Answer 2

H e y a !

Here is the solution !

Factorize -:

(p-q)^3+(q-r)^3+(r-p)^3

If a+b+c = 0 ,

then a³+b³+c³ = 3abc...

Here ,
✔a = p-q
✔b = q-r
✔c = r-p

a+b+c => (p-q)+(q-r)+(r-p)
p-q+q-r+r-p = 0

Hence , (p-q)³+(q-r)³+(r-p)³ = 3(p-q)(q-r)(r-p)

#hope it helps !