Question

if x=7+4 root 3 find the value of (x³+1/x³).​

Answer 1

Answer:

x= 11³

x= 537+1/11

x = 538/11

x= 48

Answer 2

$x \: = \: 7 \: + \: 4 \sqrt{3}$

________ [GIVEN]

${x}^{3} \: + \: \dfrac{1}{ {x}^{3} }$

________ [TO FIND]

Solution:

=> x = 7 + 4√3 _________(eq 1)

=> $\dfrac{1}{x}$ = $\dfrac{1}{ 7 \: + \: 4 \sqrt{3} }$

• Rationalize

=> $\dfrac{1}{x}$ = $\dfrac{1}{ 7 \: + \: 4 \sqrt{3} }$ × $\dfrac{7 \: - \: 4 \sqrt{3} }{ 7 \: - \: 4 \sqrt{3} }$

=> $\dfrac{7 \: - \: 4 \sqrt{3} }{ ( {7)}^{2} \: - {(4 \sqrt{3}) }^{2} }$

=> $\dfrac{7 \: - \: 4 \sqrt{3} }{49 \: - \: 48}$

=> $\dfrac{7 \: - \: 4 \sqrt{3} }{1}$

=> 7 - 4√3 ________(eq 2)

_____________________________

Now we know the value of x + $\dfrac{1}{x}$

=> x + $\dfrac{1}{x}$ = 7 + 4√3 + 7 - 4√3

=> x + $\dfrac{1}{x}$ = 14

• Now take cube on both sides

=> ${(x \: + \: \dfrac{1}{x})}^{3}$ = (14)³

(a + b)³ = a³ + 3ab(a + b) + b³

=> x³ + $\dfrac{1}{ {x}^{3} }$ + 3( $\dfrac{x}{x})$(x + $\dfrac{1}{x}$) = 2744

=> x³ + $\dfrac{1}{ {x}^{3} }$ + 3(14) = 2744

=> x³ + $\dfrac{1}{ {x}^{3} }$ = 2744 - 42

______________________________

${x}^{3} \: + \: \dfrac{1}{ {x}^{3} }$ = 2702

_____________ [ANSWER]