Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Without finding the cubes, factorize (x-2y)^{3}+(2y-3z)^{3}+(3z-x)^{3}

Without finding the cubes, factorize  (x-2y)^{3}+(2y-3z)^{3}+(3z-x)^{3}

Answers (1)

3(x-2y)(2y-3z)(3z-x)

solution :

given  (x-2y)^{3}+(2y-3z)^{3}+(3z-x)^{3}

We know that

a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)

When a + b + c = 0, we get a^{3}+b^{3}+c^{3}=3abc

Here,

(x-2y)+(2y-3z)+(3z-x)=0

So using the above identity, we get:

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

=3(x-2y)(2y-3z)(3z-x)

Hence the answer is 3(x-2y)(2y-3z)(3z-x).

Posted by

infoexpert24

View full answer

Similar Questions

Latest Question