Question

pls answer thus urgently

Answer 1

Given question,

(2-a)³+ (2 - b)³ + (2 - c)³ - 3(2 - a)(2 - b)(2 - c)


Using the identity,

x³ + y³:+ z³ - 3xyz =
(x + y + z)(x² + y² + z² - xy - yz - xz)

Assuming, (2 - a) as x
(2 - b) as y and (2 - c) as z

We get,

(2 - a)³ + (2 - b)³ + (2 - c)³ - 3(2 - a)(2 - b)(2 - c)

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

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

since a + b + c = 6, we can write,

(6 - 6)[ (2 - a)² + (2 - b)² + (2 - c)² - (2 - a)(2 - b) - (2 - b)(2 - c) - (2 - a)(2 - c) ]

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

= 0

Since 0 is multiplied, whole of the product will be equal to 0

Hence your answer is 0

Answer 2

hi there

Using the identity, x³ + y³+ z³ - 3xyz = ( x + y + z )( x² + y² + z² - xy - yz - zx )

let us take ( 2 - a ) = x (2 - b ) = y and ( 2 - c ) = z

We get, ( 2 - a )³ + ( 2 - b )³ + ( 2 - c )³ - 3( 2 - a )( 2 - b )( 2 - c )

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

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

(6 - 6) [ ( 2 - a )² + ( 2 - b )² + ( 2- c )² - ( 2 - a )(2 - b) - (2 - b)(2 - c) - ( 2 - a )( 2 - c )

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

0 multiplied by any number te product will also be 0

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