3

I am stuck with the following problem which one of friends gave me :

If $\,x-\frac 1 x=k, \, k$ being any integer,then $\,\,x^5-\frac {1}{x^5}=?$

The options are $\,\,k^5+4k^3+4k, \,k^5+5k^3+6k,\,k^5+5k^3+5k,\,k^5+5k^3+4k $.

We see that $x-\frac 1 x=k \implies x=\frac{k \pm \sqrt{k^2+4}}{2}$. Now putting this value to $\,\,x^5-\frac {1}{x^2}$ makes the calculation complicated.

Can anyone help? Thanks and regards to all.

EDIT: The problem contained a typo and thanks to @noam for pointing that out. Now using binomial expansion of $x^5-\frac{1}{x^5}$, we see that option 3 is the correct choice.

Community
  • 1
learner
  • 6,726
  • Must be a typo in the problem: try $x^5 - \frac1{x^5}$ (denominator has exponent $5$, not $2$). – Noam D. Elkies Sep 13 '13 at 13:44
  • Yet nother way $$k^2=x^2+\frac1{x^2}-2\iff k^2+2=x^2+\frac1{x^2}$$ $$k^3=x^3-\frac1{x^3}-3(x-\frac1x)\iff k^3+3k=x^3-\frac1{x^3}$$ $$(k^2+2)(k^3+3k)=x^5+(x-\frac1x)-\frac1{x^5}=x^5-\frac1{x^5}+k$$ $$k^5+5 k^3+5k=x^5-\frac1{x^5}$$ – chubakueno Sep 14 '13 at 16:35

3 Answers3

3

Let $a=x,b=\frac 1x$ and hence $$a^5-b^5=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)$$ where $a-b=k,a^2+b^2=k^2+2,\,a^4+b^4=k^4+4k^2+2$. Now putting this values in the expression ,we get $a^5-b^5=k^5+5k^3+5k$.

learner
  • 6,726
1

HINT: Applying the binomial expansion $$ k^5=\left(x-\frac1x\right)^5-5\left(x^3-\frac1{x^3}\right)+10k. $$ Now expand $k^3=\left(x-\frac1x\right)^3$ to express the cubic factor in terms of $k$.

Andrea Mori
  • 26,969
1

A little bit of cheating would be to plug in couple of values of k and see which answer works. Once you find the correct answer, one should actually still try to confirm it algebraically.

imranfat
  • 10,029