Question

Prove that 2x^3+2y^3+2z^3-6xyz=((x+y+z){(x-y)^2+(y-z)^2+(z-x)^2} hence evaluate 2(13)^3+2(14)^3+2(15)^3-6*13*14*15. Plz give the answer in detail.....Plz solve this........The best answer will be marked as brainliest.

Answer 1

Hey Mate !

Here is your solution :

R.H.S = ( x + y + z ) { ( x - y )² + ( y - z )² + ( z - x )² }.

L.H.S = 2x³ + 2y³ + 2z³ - 6xyz.

By taking out 2 as common ,

= 2 ( x³ + y³ + z³ - 3xyz )


= 2 ( x + y + z ) ( x² + y² + z² - xy - yz - zx )

= ( x + y + z ) ( 2x² + 2y² + 2z² - 2xy - 2yz - 2zx )

= ( x + y + z ) { ( x - y )² + ( y - z )² + ( z - x )²}

Proved !!

Identities used in this solution :

★ a³ + b³ + c³ - 3ab

= ( a + b + c ) ( a² + b² + c² - ab - bc - ca )

★ 2x² + 2y² + 2z² - 2xy - 2yz - 2zx

= ( x - y )² + ( y - z )² + ( z - x )²

Proof :

= 2x² + 2y² + 2z² - 2xy - 2yz - 2zx

Rearranging the terms :

= x² + y² - 2xy + y² + z² - 2yz + x² + z² - 2zx


Using identity :

[ ( a² + b² - 2ab ) = ( a - b )² ]

= ( x - y )² + ( y - z )² + ( z - x )²

Proved !!

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= 2( 13 )³ + 2( 14 )³ + 2( 15 )³ - 6 × 13 × 14 ×15

Suppose ,

13 = a , 14 = b and 15 = c

Now,

= 2a³ + 2b³ + 2c³ - 6abc

Using identity :

= ( a + b + c ) { ( a - b )²+ ( b - c )²+ ( c - a )²}

By putting the value of a,b and c.

= ( 13 + 14 + 15 ) { ( 13 - 14 )² + ( 14 - 15 )² + ( 15 - 13 )² }


= ( 42 ) { ( -1 )² + ( -1 )² + ( 2 )² }

= 42 ( 1 + 1 + 4 )

= 42 × 6

= 252

The required answer is 252.

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Hope it helps !! ^_^

Answer 2

Answer:

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Step-by-step explanation:

252