when a+b+c=x/2,then the value of (x/3 - 2a)³+(x/3- 2b)³+(x/3 - 2c)³-3(x/3 -2a)(x/3-2b)(x/3-2c)

Answer:

Given a + b + c = $\frac{x}{2}$
Then value of

($\frac{x}{3}$ -2a )³ + ($\frac{x}{3}$-2b )³ + ( $\frac{x}{3}$ - 2c )³ - 3( $\frac{x}{3}$ -2a ) ( $\frac{x}{3}$ - 2b ) ( $\frac{x}{3}$ - 2c )
We know

x³ + y³ + z³ - 3xyz  = ( x + y + z ) ( x² + y² + z^​2  - xy - yz - zx  )

So we can see that here
x = ​($\frac{x}{3}$ -2a )

y = ​($\frac{x}{3}$ -2b )

z = ​($\frac{x}{3}$ -2c )

Now , we apply formula nad get

$\Rightarrow$[ ​($\frac{x}{3}$ -2a ) + ​($\frac{x}{3}$ -2b ) + ​($\frac{x}{3}$ -2a ) ] [ ​($\frac{x}{3}$ -2a )² + ​($\frac{x}{3}$ -2b )² + ​($\frac{x}{3}$ -2c )² - ​($\frac{x}{3}$ -2a )​($\frac{x}{3}$ -2b ) - ​($\frac{x}{3}$ -2b )​($\frac{x}{3}$ -2c ) - ​($\frac{x}{3}$ -2c )​($\frac{x}{3}$ -2a ) ]

Let [ ​($\frac{x}{3}$ -2a )² + ​($\frac{x}{3}$ -2b )² + ​($\frac{x}{3}$ -2c )² - ​($\frac{x}{3}$ -2a )​($\frac{x}{3}$ -2b ) - ​($\frac{x}{3}$ -2b )​($\frac{x}{3}$ -2c ) - ​($\frac{x}{3}$ -2c )​($\frac{x}{3}$ -2a ) ]​ = A

So,

$\Rightarrow$[ ​($\frac{x}{3}$ -2a ) + ​($\frac{x}{3}$ -2b ) + ​($\frac{x}{3}$ -2a ) ] A

$\Rightarrow$[ ​$\frac{x}{3}$+ $\frac{x}{3}$ + $\frac{x}{3}$  -2a -2b -2a ] A

$\Rightarrow$[ x -2 ( a + b + c ) ] A

Now As given a + b + c = $\frac{x}{2}$ , we get
$\Rightarrow$[ x -2 ( $\frac{x}{2}$ )] A

$\Rightarrow$ [ x - x ] A

$\Rightarrow$ 0 $\times$ A

$\Rightarrow$ 0
So,

($\frac{x}{3}$ -2a )³ + ($\frac{x}{3}$-2b )³ + ( $\frac{x}{3}$ - 2c )³ - 3( $\frac{x}{3}$ -2a ) ( $\frac{x}{3}$ - 2b ) ( $\frac{x}{3}$ - 2c )  = 0             ( Ans )