Question

please anyone help me with this question
It has 100 points​

Answer 1

Answer:we know that,

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

If a+b+c=0 then,

a³+b³+c³-3abc=0×(a²+b²+c²-ab-bc-ca)=0 .....(1)

(x-a)³+(x-b)³+(x-c)³-3(x-a)(x-b)(x-c)

let a=(x-a)

b=(x-b)

and c=(x-c)

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

(x-a)³+(x-b)³+(x-c)³-3(x-a)(x-b)(x-c)=0

Thus the answer is 0....

Step-by-step explanation:

Answer 2

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.

Solution : -

Let us Assume That :-

⇒ ( x - a ) = A

⇒ ( x - b ) = B

⇒ ( x - c ) = C

Now, Using Algebric identity, we get,

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

→ A³ + B³ + C³ - 3ABC = ( A + B + C ) ( A² + B² + C² - AB - BC - CA )

Substitute back the values of A , B and C we get,

→ ( 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 , we get,

→ ( 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 (Ans).