Question

X cube + y cube + z cube - 3xyz = 1/2(x +y + z)[(x - y) whole square + (y - z) whole square + (z - x ) whole square]

Answer 1

Answer:

LHS = RHS (proved )

Step-by-step explanation:

LHS :

${x}^{3} + {y}^{3} + {z}^{3} - 3xyz \\ \\ = (x + y + z)( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - xz)$

RHS:

$\frac{1}{2} (x + y + z)( {(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2} \\ \\ = \frac{1}{2} (x + y + z)( {x}^{2} - 2xy + {y}^{2} + {y}^{2} - 2yz + {z}^{2} + {z}^{2} - 2xz + {x}^{2} ) \\ \\ = > \frac{1}{2} (x + y + z)(2 {x}^{2} + 2 {y}^{2} +2 {z}^{2} - 2xy - 2xz - 2yz) \\ \\ > \frac{1}{2} (x + y + z)2( {x}^{2} + {y}^{2} + {z}^{2} - xy - xz - yz) \\ \\ = > (x + y + z)( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - xz)$

Formula used :

${a}^{3} + {b}^{3} + {c}^{3} - 3abc \\ = > (a + b + c) ( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca)$

${(a - b)}^{2} = {a}^{2} - 2ab + {b}^{2}$

Answer 2

Answer: BY LHS AND RHS WE CAN PROVE IT..

Step-by-step explanation:

LHS:

{x}^{3}  +  {y}^{3}  +  {z}^{3}  - 3xyz

= (x + y + z)( {x}^{2}  +  {y}^{2}  +  {z}^{2}  - xy - yz - xz)

RHS:

{1}{2} (x + y + z)( {(x  -  y)}^{2}  +  {(y - z)}^{2}  +  {(z - x)}^{2}

={1}{2} (x + y + z)( {x}^{2}  - 2xy +  {y}^{2} +   {y}^{2}  - 2yz +  {z}^{2}  +  {z}^{2}  - 2xz +  {x}^{2} )

={1}{2} (x + y + z)(2 {x}^{2}  + 2 {y}^{2}  +2  {z}^{2}  - 2xy - 2xz - 2yz) {1}{2} (x + y + z)2( {x}^{2}  +  {y}^{2}  +  {z}^{2}  - xy - xz - yz)=(x + y + z)( {x}^{2}  +  {y}^{2}  +  {z}^{2}  - xy - yz - xz)

Formula used:

{a}^{3}  +  {b}^{3}  +  {c}^{3} - 3abc=(a + b + c)

( {a}^{2}  +  {b}^{2}  +  {c}^{2}  - ab - bc  - ca)  

{(a - b)}^{2}  =  {a}^{2}  - 2ab +  {b}^{2}

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