Question

x3 +1/x3 = 110,then x+1/x =

Answer 1

x³ + 1/x³  = 110

(x + 1/x)³ - 3(x + 1/x) = 110

x + 1/x = k let

k³ - 3k - 110 = 0 

k(k² - 3) = 110 

k(k² - 3) = 5(5² - 3)

so k = 5 ANSWER

Answer 2

Answer:

The value of $\frac { x + 1 } { x } = 5$

To find:

$\frac { \mathrm { x } + 1 } { \mathrm { x } } = ?$

Solution:

Given that

$\mathrm { x } ^ { 3 } + \frac { 1 } { \mathrm { x } ^ { 3 } } = 110$

Above equation can be written as like

$\left( \frac { \mathrm { x } + 1 } { \mathrm { x } } \right) ^ { 3 } - 3 \left( \frac { \mathrm { x } + 1 } { \mathrm { x } } \right) = 110$

Assume $\frac { x + 1 } { x } = k$

$\begin{array} { l } { \mathrm { k } ^ { 3 } - 3 \mathrm { k } = 110 } \\\\ { \mathrm { k } ^ { 3 } - 3 \mathrm { k } - 110 = 0 } \\\\ { \mathrm { k } \left( \mathrm { k } ^ { 2 } - 3 \right) = 110 } \end{array}$

Put k=5, we get

$\begin{array} { l } { 5 \left( 5 ^ { 2 } - 3 \right) = 110 } \\\\ { 5 ( 25 - 3 ) = 110 } \\\\ { 5 \times 22 = 110 } \\\\ { 110 = 110 } \end{array}$

Hence (k-5) is a factor

By dividing $(k^3-3k-110)$ with (k- 5) we will get the quotient as $(k^2+5k+22)$

$\mathrm { k } ^ { 3 } - 3 \mathrm { k } - 110 = ( \mathrm { k } - 5 ) \left( \mathrm { k } ^ { 2 } + 5 \mathrm { k } + 22 \right)$

Hence $( \mathrm { k } - 5 ) = 0 \text { and } \left( \mathrm { k } ^ { 2 } + 5 \mathrm { k } + 22 \right) \neq 0$

$\begin{array} { l } { \left( \frac { \mathrm { x } + 1 } { \mathrm { x } } \right) - 5 = 0 } \\\\ { \frac { \mathrm { x } + 1 } { \mathrm { x } } = 5 } \end{array}$

Therefore the value of

$\frac { \mathrm { x } + 1 } { \mathrm { x } } = 5$