Math, asked by advamitdarekar, 1 year ago

pls answer this question for expansion lesson ​

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Answers

Answered by BrainlyPopularman
4

Question :

• If { \bold{ \:  \:  {x}^{2}  +  \dfrac{1}{ 9{x}^{2} } =   \dfrac{46}{3} \:  \:  }} , Then find { \bold{ \:  \:  {x}^{3}  +  \dfrac{1}{ 27{x}^{3} } = ? \:  \:  }}

ANSWER :

 \\ \implies { \bold{ \:  \:  {x}^{2}  +  \dfrac{1}{ 9{x}^{2} } =   \dfrac{46}{3} \:  \:  }} \\

• We should write this as –

 \\ \implies { \bold{ \:  \:  {x}^{2}  +  \dfrac{1}{ {(3x)}^{2} } =   \dfrac{46}{3} \:  \:  }} \\

• Now Let's Using the identity –

 \\ \ { \huge{. \:  \:  \: }} \large {  \green{ \boxed{\bold{  {(a + b)}^{2}   =  {a}^{2}  +  {b}^{2}  + 2ab  }}}} \\

• So that –

 \\ \implies { \bold{ \:  \:  {(x +  \frac{1}{3x} )}^{2}   = {x}^{2}  +  \dfrac{1}{ {(3x)}^{2} } + 2( \cancel x)( \dfrac{1}{3 \cancel x}) \:  \:  }} \\

 \\ \implies { \bold{ \:  \:  {(x +  \frac{1}{3x} )}^{2}   =  \dfrac{46}{3} + \dfrac{2}{3} \:  \:  }} \\

 \\ \implies { \bold{ \:  \:  {(x +  \frac{1}{3x} )}^{2}   =   \cancel\dfrac{48}{3}  \:  \:  }} \\

 \\ \implies { \bold{ \:  \:  {(x +  \frac{1}{3x} )}^{2}   =  16  \:  \:  }} \\

 \\ \implies { \pink{ \bold{ \:  \:  x +  \frac{1}{3x}   =   \pm4  \:  \:  ----eq.(1)}}} \\

 \\  \:  \: \large{ \red{  \star \:  \:   \:  { \bold{ \underline{When \:  \: x +  \frac{1}{3x}  = 4} :  - }}}} \\

• Now take cube on both side –

 \\ \implies { \bold{ \:  \:  ({x +  \frac{1}{3x} }) ^{3}   =   (4)^{3}   \:  \:  }} \\

• We know that –

 \\ \ { \huge{. \:  \:  \: }} \large {  \green{ \boxed{\bold{  {(a + b)}^{3}   =  {a}^{3}  +  {b}^{3}  + 3ab(a + b)  }}}} \\

• So that –

 \\ \implies { \bold{ \:  \:   {x}^{3} +  \frac{1}{27 {x}^{3} }  +  \cancel3( \cancel x)(  \frac{1}{ \cancel{3x}} ) ({x +  \frac{1}{3x} })   =   64   \:  \:  }} \\

 \\ \implies { \bold{ \:  \:   {x}^{3} +  \frac{1}{27 {x}^{3} } +(4)   =   64   \:  \:  }} \\

 \\ \implies { \blue{ \boxed{ \bold{ \:  \:   {x}^{3} +  \frac{1}{27 {x}^{3} }  =   60  \:  \:  }}}} \\

 \\  \:  \: \large{ \red{  \star \:  \:   \:  { \bold{ \underline{When \:  \: x +  \frac{1}{3x}  =  - 4} :  - }}}} \\

• Now take cube on both side –

 \\ \implies { \bold{ \:  \:  ({x +  \frac{1}{3x} }) ^{3}   =   ( - 4)^{3}   \:  \:  }} \\

 \\ \implies { \bold{ \:  \:   {x}^{3} +  \frac{1}{27 {x}^{3} }  + ( - 4)   =    -  64   \:  \:  }} \\

 \\ \implies { \bold{ \:  \:   {x}^{3} +  \frac{1}{27 {x}^{3} }   =    -  64   + 4 \:  \:  }} \\

 \\ \implies { \blue{ \boxed{ \bold{ \:  \:   {x}^{3} +  \frac{1}{27 {x}^{3} }  =  - 60  \:  \:  }}}} \\

Hence,  \large { \red { \boxed { \bold{ \:  \:  {x}^{3}  +  \dfrac{1}{ 27{x}^{3} } = \pm 60 \:  \:  }}}} \\

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