Question

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

Answer 1

Given Equation is a + b + c = 3x.

Given (x - a)^3 + (x - b)^3 + (x - c)^3 - 3(x - a)(x - b)(x - c)

Given Equation is in the form of a^3 + b^3 + c^3 - 3abc.

We know that a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)

= > (x - a + x - b + x - c)[(x - a)^2 + (x - b)^2 + (x - c)^2 - (x - a)(x - b) - (x - b)(x - a) - (x - c)(x - a)]

= > 3x - (a + b + c)[(x - a)^2 + (x - b)^2 + (x - c)^2 - (x - a)(x - b) - (x - b)(x - a) - (x - c)(x - a)]

= > 3x - 3x

= > 0.



Hope this helps!

Answer 2

Q. 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.

Solution :

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

Let,

⇒ ( x - a ) = A

⇒ ( x - b ) = B

⇒ ( x - c ) = C

Now,

= A³ + B³ + C³ - 3ABC

Using Algebric identity,

= ( a³ + b³ + c³ - 3abc ) = ( a + b + c ) ( a² + b² + c² - ab - bc - ca )

= ( A + B + C ) ( A² + B² + C² - AB - BC - CA )

Substitute back the values of A , B and C.

= ( x - a + x - b + x - c ) [ ( x - a )² + ( x - b )² + ( x - c )² - ( x - a ) ( x - b ) - ( x - b ) ( x - c ) - ( x - a ) ( x - c ) ]

= [ 3x - ( a + b + c ) ] [ ( x - a )² + ( x - b )² + ( x - c )² - ( x - a ) ( x - b ) - ( x - b ) ( x - c ) - ( x - a ) ( x - c ) ]

Substitute the given value of ( a + b + c ),

= ( 3x - 3x ) [ ( x - a )² + ( x - b )² + ( x - c )² - ( x - a ) ( x - b ) - ( x - b ) ( x - c ) - ( x - a ) ( x - c ) ]

= ( 0 ) [ ( x - a )² + ( x - b )² + ( x - c )² - ( x - a ) ( x - b ) - ( x - b ) ( x - c ) - ( x - a ) ( x - c ) ]

= 0