Question

simplify (2x+y-z)^2-(2x-y+z)^2

Answer 1

To simplify -

Simplify ( 2x + y - z )² - ( 2x - y + z )² .

Solution -

We know that -

( a + b - c )² = a² + b² + c² + 2ab - 2ac - 2bc .

( a - b + c )² = a² + b² + c² - 2ab - 2bc + 2ac .

Using these identities here -

( 2x + y - z )²

=> [ 2x ]² + [ y ]² + [ z ] ² + 2 ( 2x )( y ) + 2 ( y )( -z ) + 2 ( 2x )( -z ) .

=> 4x² + y² + z² + 4xy - 2yz - 4xz . ...... { 1 }

( 2x - y + z )²

=> [ 2x ]² + [ y ]² + [ z ] ² + 2 ( 2x )( -y ) + 2 (- y )( z ) + 2 ( 2x )( z ) .

=> 4x² + y² + z² - 4xy - 2yz + 4xz . ...... { 2 }

[ 1 ] - [ 2 ]

→ 4x² + y² + z² + 4xy - 2yz - 4xz - ( 4x² + y² + z² - 4xy - 2yz + 4xz )

→ 4x² + y² + z² + 4xy - 2yz - 4xz - 4x² - y² - z² + 4xy + 2yz - 4xz )

→ 4xy - 2yz - 4xz - 4xy + 2yz - 4xz

→ -8 xZ .

This is the required answer.

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Answer 2

Given:-

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$\bf \implies(2x + y - z {)}^{2} - (2x - y + z {)}^{2}$

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To Find:-

⠀

$\bf \implies the \: solution$

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STEP BY STEP EXPLANATION:-

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$\bf \implies(2x + y - z {)}^{2} - (2x - y + z {)}^{2} \\ \\ \\ \bf \implies4 {x}^{2} + {y}^{2} + {z}^{2} + 4xy - 4xz - 2yz - (2y - y + z {)}^{2} \\ \\ \\ \bf \implies4 {x}^{2} + {y}^{2} + {z}^{2} + 4xy - 4xz - 2yz - (y + z {)}^{2} \\ \\ \\ \bf \implies4 {x}^{2} + {y}^{2} + {z}^{2} + 4xy - 4xz - 2yz - ( {y}^{2} + 2y z+ {z}^{2} ) \\ \\ \\ \bf \implies4 {x}^{2} + {y}^{2} + {z}^{2} + 4xy - 4xz - 2yz - {y}^{2} - 2y z - {z}^{2} \\ \\ \\ \bf \implies4 {x}^{2} + {z}^{2} + 4xy - 4xz - 2yz - {z}^{2} \\ \\ \\ \bf \implies4 {x}^{2} + 4xy - 4xz - 2yz - 2yz \\ \\ \\ \bf \implies4 {x}^{2} + 4xy - 4xz - 4yz$