# Find the value of the question below [closed]

If $x^{3}+\frac{1}{x^{3}}=14$

Find the value of $$x^{6}+\frac{1}{x^{6}}$$

Original Question:

If $x^{2}+\frac{1}{x^{2}}=14$

Find the value of $$x^{5}+\frac{1}{x^{5}}$$

• I know the answer, just sharing a new information Aug 3, 2015 at 2:05
• @Dr.MV thank you very much. I know how to improve it. thanks Aug 3, 2015 at 3:15
• This question has been edited three times, each time invalidating the previously posted answers. OP seems eager to share a lot of new information with us.
– Ian
Aug 3, 2015 at 4:57
• @OP: It is considered inappropriate behavior on this site to "recycle" questions. Continuing such actions is likely grounds for suspension. Aug 3, 2015 at 5:00
• In case you need a confirmation, here's the word from your friendly moderator. DO NOT REPLACE AN EXISTING QUESTION WITH ANOTHER ONE. This is considered rude to the people who answered the original version, because you make their posts look out of place. Remember, the answers are not just for you - they are (mostly ?) for the benefit of all the future readers of this thread. Aug 3, 2015 at 6:40

We have \begin{align*} x^5+\frac{1}{x^5}&=\left(x+\frac{1}{x}\right)\left(x^4-x^2+1-\frac{1}{x^2}+\frac{1}{x^4}\right)\\ &=\left(x+\frac{1}{x}\right)\left[\left(x^4+2+\frac{1}{x^4}\right)-\left(x^2+2+\frac{1}{x^2}\right)+1\right]\\ &=\left(x+\frac{1}{x}\right)\left[\left(14\right)^2-\left(16\right)+1\right]\\ &=181\left(x+\frac{1}{x}\right) \end{align*} Also, \begin{align*} \left|x+\frac{1}{x}\right|&=\sqrt{\left(x+\frac{1}{x}\right)^2}=\sqrt{x^2+2+\frac{1}{x^2}}=\sqrt{16}=4 \end{align*} Then $$x^5+\frac{1}{x^5}=-724\text{ or }724$$

On the other hand, $$x^3+\frac{1}{x^3}=14\implies x^6+\frac{1}{x^6}=\left(x^3+\frac{1}{x^3}\right)^2-2=(14)^2-2=194$$

• Well done!! A +1 Aug 3, 2015 at 2:16
• I knew you could just push it through by multiplying by $x^2$ and then substituting what you got into the second equation, but this is the elegant solution to be desired. Aug 3, 2015 at 2:17
• For questions of this type the key is to observe that the truly interesting variable is $x+\dfrac1x$ - not $x$ itself. +1 for making that observation. Aug 3, 2015 at 6:45

Multiply the entire expression by $x^2$, and you get $x^4 +1=14 x^2$. Next, substitute $u=x^2$, and you find $u^2-14 u +1 = 0$. Use the quadratic formula to solve for $u$, and you find $u=\frac{14\pm\sqrt{14^2 -4}}{2}$. Use this to solve for $x$. (Don't forget that x may have either sign here, since $u=x^2$)

• ah, thank you. (corrected)
– Bob
Aug 3, 2015 at 2:04
• The brute force is just as effective here as others. +1 Aug 3, 2015 at 2:17
• Just like in the NFL. That and I have a tricky (research-related) question that has been bothering me for a few weeks. I need to up my rep to put a good bounty on it. =P
– Bob
Aug 3, 2015 at 2:26

You can think of this in terms of symmetric polynomials. Let $y=1/x$ then you have:

$$x^2+y^2=14\\xy=1$$

So $$(x+y)^2=14+2\cdot 1=16$$ or $$x+y=\pm 4.$$

Now, \begin{align}x^5+y^5&=(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)\\ &=(x+y)(x^4+y^4+(xy)^2 - xy(x^2+y^2))\\ &=(x+y)\left((x^2+y^2)^2 - x^2y^2-xy(x^2+y^2)\right) \end{align} Using $x^2+y^2=14$ and $xy=1$ we have:

$$x^5+y^5=(x+y)(14^2-1-14)=\pm 4\cdot 181$$

• Very nice! I love how this method can be so easily generalized to other problems! Aug 3, 2015 at 2:40

For the newer problem, the OP asks if $x^3+x^{-3}=14$, then what is $x^6+x^{-6}$. This one is a bit easier than the earlier post.

Note that we have

$$\left(x^3+x^{-3}\right)^2=x^6+x^{-6}+2\implies x^6+x^{-6}=(14)^2-2=194$$

and we're done.

• As the OP is changed from time to time... $$x^p + 1/x^p = 14 \Rightarrow x = \sqrt[p]{ - 7 \pm 4 \sqrt{3} }$$ So $$x^q + 1/x^q = \big( - 7 \pm 4 \sqrt{3} \big)^{q/p} + 1 / \big( - 7 \pm 4 \sqrt{3} \big)^{q/p}$$ Aug 3, 2015 at 7:47