# If x + 1/x = 11 , then find the value of x^4 + 1/x^4 ?

Jun 9, 2018

${x}^{4} + \frac{1}{x} ^ 4 = 14159$

#### Explanation:

As per the question, we have

$x + \frac{1}{x} = 11$

$\therefore {\left(x + \frac{1}{x}\right)}^{2} = {\left(11\right)}^{2}$ ... [Squaring both sides]

$\therefore {x}^{2} + \frac{1}{x} ^ 2 + 2 \left(x\right) \left(\frac{1}{x}\right) = 121$

$\therefore {x}^{2} + \frac{1}{x} ^ 2 + 2 \left(\cancel{x}\right) \left(\frac{1}{\cancel{x}}\right) = 121$

$\therefore {x}^{2} + \frac{1}{x} ^ 2 + 2 = 121$

$\therefore {x}^{2} + \frac{1}{x} ^ 2 = 121 - 2 = 119$ ... (i)

Now, back to $x + \frac{1}{x} = 11$

${\left(x + \frac{1}{x}\right)}^{4} = {\left(11\right)}^{4}$

$\therefore {x}^{4} + \frac{1}{x} ^ 4 + 4 \left({x}^{3}\right) \left(\frac{1}{x}\right) + 6 \left({x}^{2}\right) \left(\frac{1}{x} ^ 2\right) + 4 \left(x\right) \left(\frac{1}{x} ^ 3\right) = {\left(11\right)}^{4}$

$\therefore {x}^{4} + \frac{1}{x} ^ 4 + 4 \left({x}^{2}\right) + 6 + 4 \left(\frac{1}{x} ^ 2\right) = 14641$

$\therefore {x}^{4} + \frac{1}{x} ^ 4 + 4 \left({x}^{2}\right) + 4 \left(\frac{1}{x} ^ 2\right) = 14641 - 6$

$\therefore {x}^{4} + \frac{1}{x} ^ 4 + 4 \left({x}^{2} + \frac{1}{x} ^ 2\right) = 14635$

$\therefore {x}^{4} + \frac{1}{x} ^ 4 + 4 \left(119\right) = 14635$ ... [Substituting the value of ${x}^{2} + \frac{1}{x} ^ 2$ from (i)]

$\therefore {x}^{4} + \frac{1}{x} ^ 4 + 476 = 14635$

$\therefore {x}^{4} + \frac{1}{x} ^ 4 = 14635 - 476$

$\therefore {x}^{4} + \frac{1}{x} ^ 4 = 14159$

Note: If you want to know more about the identity ${\left(a + b\right)}^{4}$ or any other, go to THIS SITE.

Jun 9, 2018

$14159$

#### Explanation:

We use that
${\left(a + b\right)}^{4} = {a}^{4} + 4 {a}^{3} b + 6 {a}^{2} {b}^{2} + 4 a {b}^{3} + {b}^{4}$
so we get
${\left(x + \frac{1}{x}\right)}^{4} = {x}^{4} + \frac{1}{x} ^ 4 + 4 \left({x}^{2} + \frac{1}{x} ^ 2\right) = {11}^{4} - 6$
and
${x}^{2} + \frac{1}{x} ^ 2 = {11}^{2} - 2$
so we get

${x}^{4} + \frac{1}{x} ^ 4 = {11}^{4} - 6 - 4 \left({11}^{2} - 2\right) = 14159$

Jun 9, 2018

${x}^{2} + \frac{1}{x} ^ 2 = {\left(x + \frac{1}{x}\right)}^{2} - 2 \cdot x \cdot \frac{1}{x}$

$\implies {x}^{2} + \frac{1}{x} ^ 2 = {11}^{2} - 2 = 119$

Now ${x}^{4} + \frac{1}{x} ^ 4 = {\left({x}^{2} + \frac{1}{x} ^ 2\right)}^{2} - 2 \cdot {x}^{2} \cdot \frac{1}{x} ^ 2$

$\implies {x}^{4} + \frac{1}{x} ^ 4 = {119}^{2} - 2 = 14159$