Question

$\huge{\large{\bf{\red{for \: {\red{\underline{\blue{\mathfrak{Maths \: Aryabhatta }}}}}}}}}$
$\huge{\large{\bf{\pink{➢{\red{\underline{\green{\mathfrak{Question:}}}}}}}}}$
QUESTION:--

$\sf\red{(x³ - y³) ÷ (x - y)}$

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Answer 1

Answer:

we have,

$\sf \dfrac{x^3-y^3}{(x-y)}$

formula,

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

$\sf \Rightarrow \dfrac{(x-y)(x^2+xy+y^2)}{(x-y)}$

$\sf \Rightarrow \dfrac{ \cancel{(x-y)}(x^2+xy+y^2)}{\cancel{(x-y)}}$

$\sf \Rightarrow x^2+xy+y^2$

Answer 2

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

Given expression is

$\rm :\longmapsto\:\dfrac{ {x}^{3} - {y}^{3} }{x - y}$

We know,

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

So, using this identity, we get

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

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

Hence,

$\red{\rm \implies\:\boxed{ \tt{ \: \dfrac{ {x}^{3} - {y}^{3} }{x - y} = {x}^{2} + xy + {y}^{2} \: }}}$

Alternative Method :-

Using Long Division

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\: {x}^{2} + xy + {y}^{2} \:\:}}}\\ {\underline{\sf{x - y}}}& {\sf{\: {x}^{3} + 0 {x}^{2}y + {0y}^{2}x -{y}^{3} \:\:}} \\{\sf{}}& \underline{\sf{\: -{x}^{3} + {yx}^{2}  \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:   \:  \:\:}} \\ {{\sf{}}}& {\sf{\: \:  {yx}^{2} + {0xy}^{2} - {y}^{3} \:    \:\:}} \\{\sf{}}& \underline{\sf{\:\: \:    - {yx}^{2} + {xy}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:   \:\:}} \\ {\underline{\sf{}}}& {\sf{\: {xy}^{2} - {y}^{3} \:\:}} \\{\sf{}}& \underline{\sf{\: \:  - {xy}^{2} + {y}^{3} \: \: \: \: \: \: \:}} \\ {\underline{\sf{}}}& {\sf{\:0\:\:}}  \end{array}\end{gathered}\end{gathered}\end{gathered}$

Hence,

$\red{\rm \implies\:\boxed{ \tt{ \: \dfrac{ {x}^{3} - {y}^{3} }{x - y} = {x}^{2} + xy + {y}^{2} \: }}}$

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Additional Information :-

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

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

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

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

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

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

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

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