Find the value of $(x-a)^3 + (x-b)^3 +
(x-c)^3 - 3 (x-a)(x-b)(x-c)$ if $a+b+c = 3x$


Given :

$a + b + c = 3x$

To find :

We have to find the value of $(x-a)^3 + (x-b)^3 + (x-c)^3 - 3 (x-a)(x-b)(x-c)$.

Solution :

We know that,

If $a + b + c = 0$ then $a^3+ b^3+ c^3=3abc$.

$(x-a)+(x-b)+(x-c)= 3x-(a+b+c)$ 

                              $= 3x-3x$

                                = 0

Therefore,

$(x-a)^3 + (x-b)^3 + (x-c)^3 - 3 (x-a)(x-b)(x-c) = 3 (x-a)(x-b)(x-c)$

                                                                     

$(x-a)^3 + (x-b)^3 + (x-c)^3 - 3 (x-a)(x-b)(x-c)  = 0$. 

The value of $(x-a)^3 + (x-b)^3 + (x-c)^3 - 3 (x-a)(x-b)(x-c)$ is 0.

Tutorialspoint
Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

33 Views

Kickstart Your Career

Get certified by completing the course

Get Started
Advertisements