Question

plz solve this question
( algebraic rational expression)
do homely
plz answer my questions first.
​

Answer 1

Qᴜᴇsᴛɪᴏɴ :-

$\bf\dfrac{y}{{y}^{2}-\dfrac{{y}^{3}-1}{y+ \dfrac{1}{y+1}}}$

Sᴏʟᴜᴛɪᴏɴ :-

Lets try to solve part by part :-

starting from Denominators,

$\purple\longmapsto\tt \: y + \dfrac{1}{y + 1} \\ \\ \purple\longmapsto\tt \: \frac{y(y + 1) + 1}{y + 1} \\ \\ \purple\longmapsto\tt \: \frac{ {y}^{2} + y + 1 }{y + 1}$

Now when we put this value in the denominator of (y³-1), numerator & denominator will change ...

we get :-

$\purple\longmapsto\tt \: \dfrac{(y^{3} - 1)(y + 1)}{( {y}^{2} + y + 1) }$

Now, putting :-

→ (y³ - 1) = (y³ - 1³) = (a³ - b³)

→ (a - b)(a² + b² + ab) = (y - 1)(y² + y + 1)

$\purple\longmapsto\tt \: \dfrac{(y - 1) \cancel{({y}^{2} + y + 1)}(y + 1)}{ \cancel{({y}^{2} + y + 1)}} \\ \\ \purple\longmapsto\tt \: (y - 1)(y + 1) \\ \\ \purple\longmapsto\tt \: ( {y}^{2} - 1)$

Now, putting this value we get,

$\purple\longmapsto\tt \: \dfrac{y}{ {y}^{2} - ( {y}^{2} - 1)} \\ \\ \purple\longmapsto\tt \: \dfrac{y}{ \cancel{{y}^{2}} - \cancel{{y}^{2}} + 1} \\ \\ \purple\longmapsto\tt \: \dfrac{y}{1} \\ \\ \purple\longmapsto\tt \bold{\boxed{\huge{\boxed{\orange{\small{\boxed{\huge{\red{\bold{\huge\bold{\red{{y}}}}}}}}}}}}}$

$\rule{200}{4}$

Answer 2

${ \sf{ \red{ \huge{ \frac{y}{ {y}^{2} - \frac{ {y}^{3} - 1 }{y + \frac{1}{y + 1}}}} } } }$




▪ First solving the expression given in the denominator of ( y^3 - 1 ) ...



${ \bold{ \purple{y + \frac{1}{y + 1}}}}$




${ \bold{ \purple{ = \frac{y(y + 1) + 1}{y + 1}}}}$



${ \bold{ \purple{ = \frac{ {y}^{2} + y + 1}{y + 1}}}}$




▪ now, putting this above expression in the denominator of ( y^3 - 1 ) ....




${ \orange{ \bold{ \frac{ {y}^{3} - 1 }{ \frac{ {y}^{2} + y + 1}{y + 1}}}} }$




${ \orange{ \bold{ = \frac{( {y}^{3} - 1)(y + 1)}{ {y}^{2} + y + 1}}} }$




we know that,




${ \boxed{ \boxed{ \sf{ \pink{ {a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2} )}}}}}$




▪ using the above given algebraic identity in the numerator side....




${ \bold{ \orange{ = \frac{(y - 1)( {y}^{2} + y + 1)(y + 1)}{( {y}^{2} + y + 1)}}}}$




${ \orange{ \bold{ = (y - 1)(y + 1)}}}$




since,




${ \boxed{ \boxed{ \sf{ \pink{(a - b)(a + b) = {a}^{2} - {b}^{2} }}}}}$




▪ then we get,




${ \orange{ \bold{ = ( {y}^{2} - 1)}}}$




▪ putting this value in the main algebraic expression



${ \bold{ \blue{ = \frac{y}{ {y}^{2} - ( {y}^{2} - 1)}}}}$




${ \blue{ \bold{ = \frac{y}{ {y}^{2} - {y}^{2} + 1}}}}$




${ \blue{ \bold{ = \frac{y}{1}}}}$



therefore,



${ \underline{ \boxed{ \boxed{ \red{ \bold{ \frac{y}{ {y}^{2} - \frac{ {y}^{3} - 1}{y + \frac{1}{y + 1} } } = y}}}}}}$