Question

pls answer this question for expansion lesson ​

Answer 1

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