Question

Prove that 2x^3 +2y^3 + 2z^3

– 6xyz =(x + y + z) [(x – y )^2 + ( y – z )^2 + ( z – x )^2 ] Give explanation step by step. ​

Answer 1

Answer:

252

Step-by-step explanation:

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 + y² + 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 !!

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

x15

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 x 6

= 252

Hence proved