# Thread: If [x+(1/x)]=3, find the value of x^5+[1/(x^5)].

1. ## If [x+(1/x)]=3, find the value of x^5+[1/(x^5)].

If (x+1/x)=3, then find the numerical value of x^5+1/x^5.

2. Is it $\displaystyle \frac{x + 1}{x} = 3$ or $\displaystyle x + \frac{1}{x} = 3$?

Are you trying to find $\displaystyle \frac{x^5 + 1}{x^5}$ or $\displaystyle x^5 + \frac{1}{x^5}$?

3. I think you will understand this better:
If [x+(1/x)]=3, then find the numerical value of x^5+[1/(x^5)].

4. Originally Posted by Kaloda
If (x+1/x)=3, then find the numerical value of x^5+1/x^5.
Things to note:

$\displaystyle \displaystyle \left(x + \frac{1}{x}\right)^2 = 9 \Rightarrow x^2 + 2 + \frac{1}{x^2} = 9 \Rightarrow x^2 + \frac{1}{x^2} = 7$ .... (A)

From (A):

$\displaystyle \displaystyle \left(x + \frac{1}{x}\right) \left(x^2 + \frac{1}{x^2}\right) = (3)(7) = 21$ .... (B)

$\displaystyle \displaystyle \left(x + \frac{1}{x}\right) \left(x^2 + \frac{1}{x^2}\right) = x^3 + x + \frac{1}{x} + \frac{1}{x^3} = \left(x^3 + \frac{1}{x^3}\right) + \left(x + \frac{1}{x}\right)$ $\displaystyle \displaystyle = x^3 + \frac{1}{x^3} + 3$ .... (C)

From (B) and (C):

$\displaystyle \displaystyle x^3 + \frac{1}{x^3} = ....$ .... (D)

etc.

It's not hard to work your way to the value of $\displaystyle \displaystyle x^5 + \frac{1}{x^5}$.

5. Thank you very very much!

6. $\displaystyle \displaystyle{{x}^{3}}+\frac{1}{{{x}^{3}}}=\left( x+\frac{1}{x} \right)\left( {{\left( x+\frac{1}{x} \right)}^{2}}-3 \right)\text{ (1)},$ and $\displaystyle \displaystyle{{x}^{4}}+\frac{1}{{{x}^{4}}}={{\left ( {{\left( x+\frac{1}{x} \right)}^{2}}-2 \right)}^{2}}-2\text{ (2)}\text{,}$ so $\displaystyle \displaystyle\left( x+\frac{1}{x} \right)\left( {{x}^{4}}+\frac{1}{{{x}^{4}}} \right)={{x}^{5}}+\frac{1}{{{x}^{3}}}+{{x}^{3}}+\f rac{1}{{{x}^{5}}},$ conclude by $\displaystyle (1)$ and $\displaystyle (2).$

7. Hello, Kaloda!

This solution should surprise/amaze/terrify your teacher . . .

$\displaystyle \text{If }x+\dfrac{1}{x} \:=\:3\text{, find the value of }x^5 + \dfrac{1}{x^5}$

We have: .$\displaystyle x + \dfrac{1}{x} \:=\:3$ .[1]

Square both sides:
. . $\displaystyle \left(x + \dfrac{1}{x}\right)^2 \:=\:3^2 \quad\Rightarrow\quad x^2 + 2 + \dfrac{1}{x^2} \:=\:9 \quad\Rightarrow\quad x^2 + \dfrac{1}{x^2} \:=\:7$

Raise [1] to the 5th power: .$\displaystyle \left(x + \dfrac{1}{x}\right)^5 \;=\;3^5$

. . . . . . . . . . . .$\displaystyle \displaystyle x^5 + 5x^3 + 10x + \frac{10}{x} + \frac{5}{x^3} + \frac{1}{x^5} \;\;=\;\;243$

. . . . . $\displaystyle \displaystyle \left(x^5 + \frac{1}{x^5}\right) + 5\underbrace{\left(x^3 + \frac{1}{x^3}\right)}_{\text{sum of cubes}} + 10\underbrace{\left(x + \frac{1}{x}\right)}_{\text{This is 3}} \;\;=\;\;243$

. . $\displaystyle \displaystyle \left(x^5 + \frac{1}{x^5}\right) + 5\underbrace{\left(x + \frac{1}{x}\right)}_{\text{This is 3}} \underbrace{\left(x^2 - 1 + \frac{1}{x^2}\right)}_{\text{This is 7 - 1}} + 30 \;\;=\;\;243$

. . . . . . . . . . . . . . . .$\displaystyle \displaystyle \left(x^5+\frac{1}{x^5}\right) + 5(3)(6) + 30 \;\;=\;\;243$

. . . . . . . . . . . . . . . . . . . Therefore: . . $\displaystyle \displaystyle x^5 + \frac{1}{x^5} \;\;=\;\;123$

8. Originally Posted by Kaloda
If (x+1/x)=3, then find the numerical value of x^5+1/x^5.
You could also use the Binomial Expansion.

$\displaystyle \displaystyle\left(x+\frac{1}{x}\right)^5=x^5+\bin om{5}{1}x^4\ \frac{1}{x}+\binom{5}{2}x^3\ \frac{1}{x^2}+\binom{5}{3}x^2\ \frac{1}{x^3}+\binom{5}{4}x\ \frac{1}{x^4}+\frac{1}{x^5}$

$\displaystyle =\displaystyle\ x^5+5x^3+10x+10\ \frac{1}{x}+5\ \frac{1}{x^3}+\frac{1}{x^5}$

$\displaystyle \displaystyle\Rightarrow\ \left(x+\frac{1}{x}\right)^5=x^5+\frac{1}{x^5}+10\ left(x+\frac{1}{x}\right)+5\left(x^3+\frac{1}{x^3} \right)$

$\displaystyle \displaystyle\Rightarrow\ x^5+\frac{1}{x^5}=\left(x+\frac{1}{x}\right)^5-10\left(x+\frac{1}{x}\right)-5\left(x^3+\frac{1}{x^3}\right)$

Also

$\displaystyle \displaystyle\left(x+\frac{1}{x}\right)^3=x^3+\bin om{3}{1}x^2\ \frac{1}{x}+\binom{3}{2}x\ \frac{1}{x^2}+\frac{1}{x^3}$

$\displaystyle =\displaystyle\ x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)$

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

Therefore

$\displaystyle \displaystyle\ x^5+\frac{1}{x^5}=\left(x+\frac{1}{x}\right)^5-10\left(x+\frac{1}{x}\right)-5\left[\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)\right]$

$\displaystyle \displaystyle\Rightarrow\ x^5+\frac{1}{x^5}=\left(x+\frac{1}{x}\right)^5-5\left(x+\frac{1}{x}\right)^3+5\left(x+\frac{1}{x} \right)$

$\displaystyle =3^5-(5)3^3+5(3)$

9. x^5 + 1/x^5 = 123

x = {[123+-SQRT(15125)] / 2}^(1/5)

x = 2.618033988.... or .381966011.... ; ya, ya, so what!

,

,

,

,

,

,

,

,

,

,

,

,

,

,

# x^5 1/x^5=

Click on a term to search for related topics.