Question

if x²-5x+1=0; find x³+1/x³​

Answer 1

Question:

If ${x}^{2} - 5x + 1 = 0$ ; find ${x}^{3} + \dfrac{1}{ {x}^{3} }$ .

To find:

★ To find the value of ${x}^{3} + \dfrac{1}{ {x}^{3} }$

Answer:

The the value of $\underline{ \:\underline{ \: \sf \pink{ {x}^{3} + \dfrac{1}{ {x}^{3} }} \: is \: \bold{ \pink{ 110}} \: }\: }$ .

Given:

★ An equation is given ${x}^{2} - 5x + 1 = 0 .$

Step-by-step explanation:

${x}^{2} - 5x + 1 = 0$

$\implies {x}^{2} + 1 = 5x$

$\implies \dfrac{ {x}^{2} + 1}{x} = 5 \: ---> ( 1 )$

Now squaring on both sides,

$\implies { \big(\dfrac{ {x}^{2} + 1}{x} \big)}^{2} = {(5)}^{2}$

$\implies {x}^{2} + \dfrac{1}{x} + 2 = 25$

$\implies {x}^{2} + \dfrac{1}{{x }^{2} } = 23 \: ---> ( 2 )$

Multiplying eq ( 1 ) and ( 2 )

$\longmapsto \big(x + \dfrac{1}{x} \big) \big( {x}^{2} + \dfrac{1}{ {x}^{2} } \big) = 5 \times 23$

$\longmapsto {x}^{3} + \dfrac{1}{ {x}^{3} } + x + \dfrac{1}{x} = 115$

$\longmapsto {x}^{3} + \dfrac{1}{ {x}^{3} } +5 = 115$

$\longmapsto \boxed{{x}^{3} + \dfrac{1}{ {x}^{3} } = 110}$

$\therefore$The the value of $\underline{ \:\underline{ \: \bold{ {x}^{3} + \dfrac{1}{ {x}^{3} }} \: is \: \bold{ 110} \: }\: }$ .

Formula used:

$\bigstar \: \tt {(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2}$