Math, asked by archanachougale9291, 1 month ago

Ifx-y=5 and xy=6find the value of x^3-y^3

Answers

Answered by mathdude500
5

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

Given that,

 \red{\rm :\longmapsto\:x - y = 5}

and

 \red{\rm :\longmapsto\:xy = 6}

Now, Consider

 \green{\rm :\longmapsto\: {x}^{3} -  {y}^{3}}

 \green{\rm \:  =  \: {(x - y)}^{3} + 3xy(x - y)}

On substituting the values, we get

 \green{\rm \:  =  \:  {5}^{3} + 3 \times 5 \times 6}

 \green{\rm \:  =  \:  125 + 90}

 \green{\rm \:  =  \:  215}

 \red{\bf\implies \:\boxed{ \tt{ \:  {x}^{3} -  {y}^{3} = 215 \: }}}

More Identities to know :-

 \green{\boxed{ \tt{ \:  {(x + y)}^{2} =  {x}^{2} + 2xy +  {y}^{2} \: }}}

 \green{\boxed{ \tt{ \:  {(x - y)}^{2} =  {x}^{2}  -  2xy +  {y}^{2} \: }}}

 \blue{\boxed{ \tt{ \:  {(x + y)}^{3} =  {x}^{3} +  {y}^{3} + 3xy(x + y) \: }}}

 \blue{\boxed{ \tt{ \:  {(x  -  y)}^{3} =  {x}^{3}  -  {y}^{3} - 3xy(x  -  y) \: }}}

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

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

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

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

 \purple{\boxed{ \tt{ \: {x}^{2} -  {y}^{2} = (x + y)(x - y)   \: }}}

 \purple{\boxed{ \tt{ \: {x}^{4} -  {y}^{4} = (x + y)(x - y)( {x}^{2} +  {y}^{2})\: }}}

Answered by BrainlyArnab
0

 \huge \red{ \boxed{ \bf \blue{215}}}

Step-by-step explanation:

GIVEN :-

 \bf \color{orange}{x - y = 5} \\  \\  \bf \color{skyblue}{xy = 6} \\

TO FIND :-

 \bf \color{lime}{ {x}^{3}  -  {y}^{3} } \\

SOLUTION :-

We know that,

 \color{maroon}{ \boxed{ \bf \color{yellow}{ {x}^{3}  -  {y}^{3} } =  \color{red}{ {(x - y)}^{3}   + 3xy(x - y)}}}

By Substituting values we get,

  =  \bf \color{green} {(5)}^{3}  + 3(6)(5) \\  \\  =  \bf \color{blue}125 + 90 \\  \\   = \large \underline{ \overline{\bf \color{purple}{215}}}

Hence,

The value of - = 215.

Hope it helps.

#BeBrainly :-)

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