Question

if x+1/x=3 find the value of x^2+1/x^2 and x^3+1/x^3​

Answer 1

$\mathfrak{\large{\underline{\underline{Answer:-}}}}$

$\bf x^2 + \dfrac{1}{x^2} = 7$

$\bf x^2 + \dfrac{1}{x^2} = 18$

$\mathfrak{\large{\underline{\underline{Explanation:-}}}}$

Given :- $\sf x + \dfrac{1}{x} = 3$

To find :- $\sf x^2 + \dfrac{1}{x^2}$

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

Solution :-

$\sf (i) \: x^2 + \dfrac{1}{x^2}$

$\tt x + \dfrac{1}{x} = 3$

By squaring on both sides

$\tt (x + \dfrac{1}{x})^{2} ={3}^{2}$

$\tt (x + \dfrac{1}{x})^{2} =3 \times 3$

$\tt (x + \dfrac{1}{x})^{2} =9$

In Left Hand Side of the above equation

We know that (a + b)² = a² + b² + 2ab

Here a = x, b = 1/x

By sustituting the values in the identity we have

$\tt {(x)}^{2} + {( \dfrac{1}{x})}^{2} + 2(x)( \dfrac{1}{x}) = 9$

$\tt {x}^{2} + {\dfrac{ {1}^{2} }{ {x}^{2} }}+ 2(x)( \dfrac{1}{x}) = 9$

$\tt {x}^{2} + {\dfrac{1}{ {x}^{2} }}+ 2(x)( \dfrac{1}{x}) = 9$

$\tt {x}^{2} + {\dfrac{1}{{x}^{2} }}+ 2( \cancel x)( \dfrac{1}{ \cancel x}) = 9$

$\tt {x}^{2} + {\dfrac{1}{{x}^{2} }}+ 2(1)= 9$

$\tt {x}^{2} + {\dfrac{1}{{x}^{2} }}+ 2= 9$

$\tt {x}^{2} + {\dfrac{1}{{x}^{2} }} = 9 - 2$

$\bf {x}^{2} + {\dfrac{1}{{x}^{2} }} = 7$

$\sf (i) \: x^3 + \dfrac{1}{x^3}$

$\tt x + \dfrac{1}{x} = 3$

By cubing on both sides

$\tt (x + \dfrac{1}{x})^{3} = {3}^{3}$

$\tt (x + \dfrac{1}{x})^{3} = 3 \times 3 \times 3$

$\tt (x + \dfrac{1}{x})^{3} = 9 \times3$

$\tt (x + \dfrac{1}{x})^{3} = 27$

In Left Hand Side of the above equation

We know that (a + b)³ = a³ + b³ + 3ab(a + b)

Here a = x, b = 1/x

By sustituting the values in the identity we have

$\tt {(x)}^{3} + {( \dfrac{1}{x})}^{3} + 3(x)( \dfrac{1}{x})(x + \dfrac{1}{x}) = 27$

$\tt {x}^{3} + { \dfrac{ {1}^{3} }{ {x}^{3} }} + 3(x)( \dfrac{1}{x})(x + \dfrac{1}{x}) = 27$

$\tt {x}^{3} + { \dfrac{1}{ {x}^{3} }} + 3( \cancel x)( \dfrac{1}{ \cancel x})(x + \dfrac{1}{x}) = 27$

$\tt {x}^{3} + { \dfrac{1}{ {x}^{3} }} + 3(1)(x + \dfrac{1}{x}) = 27$

$\tt {x}^{3} + { \dfrac{1}{ {x}^{3} }} + 3(x + \dfrac{1}{x}) = 27$

$\tt {x}^{3} + { \dfrac{1}{ {x}^{3} }} + 3(3) = 27$

[Since given that $\bf x + \dfrac{1}{x} = 3$ ]

$\tt {x}^{3} + { \dfrac{1}{ {x}^{3} }} + 9 = 27$

$\tt {x}^{3} + { \dfrac{1}{ {x}^{3} }} = 27 - 9$

$\bf {x}^{3} + { \dfrac{1}{ {x}^{3} }} = 18$

Answer 2

x + $\dfrac{1}{x}$ = 3

_________ [GIVEN]

• We have to find the value of x² + $\frac{1}{ {x}^{2} }$ and x³ + $\frac{1}{ {x}^{3} }$

______________________________

x + $\dfrac{1}{x}$ = 3

• Squaring on both sides.

=> ${(x \: + \: \dfrac{1}{x} )}^{2}$ = (3)²

(a + b)² = a² + 2ab + b²

=> x² + $\frac{1}{ {x}^{2} }$ + 2x $\dfrac{1}{x}$ = 9

=> x² + $\frac{1}{ {x}^{2} }$ + 2 = 9

=> x² + $\frac{1}{ {x}^{2} }$ = 9 - 2

______________________________

x² + $\frac{1}{ {x}^{2} }$ = 7

___________ $\bold{[ANSWER]}$

______________________________

x + $\dfrac{1}{x}$ = 3

• Cube on both sides.

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

(a + b)³ = a³ + b³ + 3ab (a + b)

=> x³ + $\frac{1}{ {x}^{3} }$ + 3x $(\dfrac{1}{x})$ $(x \: + \: \dfrac{1}{x} )$ = 27

=> x³ + $\frac{1}{ {x}^{3} }$ + 3(3) = 27

=> x³ + $\frac{1}{ {x}^{3} }$ + 9 = 27

=> x³ + $\frac{1}{ {x}^{3} }$ = 27 - 9

_____________________________

x³ + $\frac{1}{ {x}^{3} }$ = 18

____________ $\bold{[ANSWER]}$