Question

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Answer 1

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\: {13}^{3} - {5}^{3} \: is \: divisible \: by \: 8$

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

We know that

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

On substituting x = 13 and y = 8, we get

$\rm :\longmapsto\: {13}^{3} - {5}^{3}$

$\rm \: = \: \: (13 - 5)( {13}^{2} + 13 \times 5 + {5}^{2})$

$\rm \: = \: \: 8(169 + 65 + 25)$

$\rm \: = \: \: 8 \times 259$

$\bf\implies \: {13}^{3} - {5}^{3} = 8 \times 259$

$\bf\implies \: {13}^{3} - {5}^{3} = 8k$

$\bf\implies \: {13}^{3} - {5}^{3} \: is \: multiple \: of \: 8$

$\bf\implies \: {13}^{3} - {5}^{3} \: is \: divisible \: by \: 8$

Additional Information :-

More Identities to know

$\boxed{ \rm \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}}$

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

$\boxed{ \rm \: {(x + y)}^{2} - {(x - y)}^{2} = 4xy}$

$\boxed{ \rm \: {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2}) }$

$\boxed{ \rm \: (x + y)(x - y) = {x}^{2} - {y}^{2} }$

$\boxed{ \rm \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}$

$\boxed{ \rm \: {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y)}$

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

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

$\boxed{ \rm \: {x}^{4} - {y}^{4} = (x - y)(x + y)( {x}^{2} + {y}^{2})}$

Answer 2

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