Question

if x+1/x=7,then find the value of x^3+1/x^3​

Answer 1

x + $\dfrac{1}{x}$ = 7

_________ [GIVEN]

• We have to find the value of x³ + $\dfrac{1}{ {x}^{3} }$

________________________________

Solution:

=> x + $\dfrac{1}{x}$ = 7

• Take cube on both sides

=> $( {x \: + \: \dfrac{1}{x}) }^{3}$ = (7)³

=> x³ + $\dfrac{1}{ {x}^{3} }$ + 3 $(\dfrac{x}{x})$ $(x \: + \: \dfrac{1}{x})$ = 343

=> x³ + $\dfrac{1}{ {x}^{3} }$ + 3(7) = 343

=> x³ + $\dfrac{1}{ {x}^{3} }$ + 21 = 343

=> x³ + $\dfrac{1}{ {x}^{3} }$ = 343 - 21

______________________________

x³ + $\dfrac{1}{ {x}^{3} }$ = 322

___________ [ANSWER]

Answer 2

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

$\boxed{ \sf {x}^{3} + { \dfrac{1}{ {x}^{3} }} = 322}$

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

Given :- $\tt x + \dfrac{1}{x} = 7$

To find :- $\tt x^3 + \dfrac{1}{x^3}$

Solution :-

$\tt x + \dfrac{1}{x} = 7$

By cubing on both sides

$\tt (x + \dfrac{1}{x})^{3} = {7}^{3}$

$\tt (x + \dfrac{1}{x})^{3} = 7 \times 7 \times 7$

$\tt (x + \dfrac{1}{x})^{3} = 49 \times 7$

$\tt (x + \dfrac{1}{x})^{3} = 343$

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

$\tt {(x)}^{3} + {( \dfrac{1}{x})}^{3} + 3(x)( \dfrac{1}{x})(x + \dfrac{1}{x}) = 343$

$\tt {x}^{3} + { \dfrac{ {1}^{3} }{ {x}^{3} }} + 3(x)( \dfrac{1}{x})(x + \dfrac{1}{x}) = 343$

$\tt {x}^{3} + { \dfrac{1}{ {x}^{3} }} + 3( \cancel x)( \dfrac{1}{ \cancel x})(x + \dfrac{1}{x}) = 343$

$\tt {x}^{3} + { \dfrac{1}{ {x}^{3} }} + 3(1)(x + \dfrac{1}{x}) = 343$

$\tt {x}^{3} + { \dfrac{1}{ {x}^{3} }} + 3(x + \dfrac{1}{x}) = 343$

$\tt {x}^{3} + { \dfrac{1}{ {x}^{3} }} + 3(7) = 343$

[Since given that $\bf x + \dfrac{1}{x} = 7$ ]

$\tt {x}^{3} + { \dfrac{1}{ {x}^{3} }} + 21 = 343$

$\tt {x}^{3} + { \dfrac{1}{ {x}^{3} }} = 343 - 21$

$\tt {x}^{3} + { \dfrac{1}{ {x}^{3} }} = 322$

$\Huge{\boxed{ \sf {x}^{3} + { \dfrac{1}{ {x}^{3} }} = 322}}$