Question

PROVE THAT
$(x + y + z)^{3} - {x}^{3} - {y}^{3} - {z }^{3} = 3(x + y)(y + z)(z + x)$
PLS I WANT BEFORE 18:30

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Answer 1

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm :\longmapsto\: {(x + y + z)}^{3} - {x}^{3} - {y}^{3} - {z}^{3}$

can be re-arranged as

$\rm  =  [{(x + y + z)}^{3} - {x}^{3} ]-[ {y}^{3} + {z}^{3}]$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2})}}}$

and

$\purple{\rm :\longmapsto\:\boxed{\tt{ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2})}}}$

So, using these Identities, we get

$\rm = [x + y + z - x][ {(x + y + z)}^{2} + {x}^{2} + x(x + y + z)] - (y + z)( {y}^{2} - yz + {z}^{2})$

$\rm = [y + z][ {x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2zx +{x}^{2} + {x}^{2} + xy + xz] - (y + z)( {y}^{2} - yz + {z}^{2})$

$\rm = [y + z][{x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2zx +{x}^{2} + {x}^{2} + xy + xz- {y}^{2} + yz - {z}^{2}]$

$\rm = [y + z][3{x}^{2} + 3xy + 3yz + 3zx]$

$\rm = 3[y + z][{x}^{2} + xy + yz + zx]$

$\rm = 3[y + z][x(x + y)+ z(x + y)]$

$\rm = 3[y + z][(x + y)(x + z)]$

$\rm \:  =  \: 3(x + y)(y + z)(z + x)$

Hence,

$\boxed{\tt{ {(x + y + z)}^{3} - {x}^{3} - {y}^{3} - {z}^{3} = 3(x + y)(y + z)(z + x)}}$

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More Identities to know :

➢  (a + b)² = a² + 2ab + b²

➢  (a - b)² = a² - 2ab + b²

➢  a² - b² = (a + b)(a - b)

➢  (a + b)² = (a - b)² + 4ab

➢  (a - b)² = (a + b)² - 4ab

➢  (a + b)² + (a - b)² = 2(a² + b²)

➢  (a + b)³ = a³ + b³ + 3ab(a + b)

➢  (a - b)³ = a³ - b³ - 3ab(a - b)

Answer 2

Answer:

$\large\underline{\bf{Solution-}}$

Consider LHS

$\rm :\longmapsto\: {(x + y + z)}^{3} - {x}^{3} - {y}^{3} - {z}^{3}$

can be re-arranged as

$\rm  =  [{(x + y + z)}^{3} - {x}^{3} ]-[ {y}^{3} + {z}^{3}]$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2})}}}$

and

$\purple{\rm :\longmapsto\:\boxed{\tt{ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2})}}}$

So, using these Identities, we get

$\rm = [x + y + z - x][ {(x + y + z)}^{2} + {x}^{2} + x(x + y + z)] - (y + z)( {y}^{2} - yz + {z}^{2})$

$\rm = [y + z][ {x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2zx +{x}^{2} + {x}^{2} + xy + xz] - (y + z)( {y}^{2} - yz + {z}^{2})$

$\rm = [y + z][{x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2zx +{x}^{2} + {x}^{2} + xy + xz- {y}^{2} + yz - {z}^{2}]$

$\rm = [y + z][3{x}^{2} + 3xy + 3yz + 3zx]$

$\rm = 3[y + z][{x}^{2} + xy + yz + zx]$

$\rm = 3[y + z][x(x + y)+ z(x + y)]$

$\rm = 3[y + z][(x + y)(x + z)]$

$\rm \:  =  \: 3(x + y)(y + z)(z + x)$

Hence,

$\boxed{\tt{ {(x + y + z)}^{3} - {x}^{3} - {y}^{3} - {z}^{3} = 3(x + y)(y + z)(z + x)}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

More Identities to know :

➢  (a + b)² = a² + 2ab + b²

➢  (a - b)² = a² - 2ab + b²

➢  a² - b² = (a + b)(a - b)

➢  (a + b)² = (a - b)² + 4ab

➢  (a - b)² = (a + b)² - 4ab

➢  (a + b)² + (a - b)² = 2(a² + b²)

➢  (a + b)³ = a³ + b³ + 3ab(a + b)

➢  (a - b)³ = a³ - b³ - 3ab(a - b)

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