0

Given $x^4+\frac 1{x^4}=1$ Then find the value of $$x^{48}+x^{52}+x^{56}+x^{60}$$ My try ----make a quadratic equation and find x but that is too tedious. ... I tried factorisation but didn't get anything. Please help.

Vindhyachal Vindhyachal
  • 17
  • There are lots of good identities to play with here. For instance, multiplying by $x^{52}$ shows that $x^{56}+x^{48}=x^{52}$ Similarly $x^{60}+x^{52}=x^{56}$. Adding these shows that your sum equals $x^{52}+x^{56}$ which implies that $x^{48}+x^{60}=0$. Doesn't immediately solve your problem, but it seems like a promising start. – lulu Aug 09 '17 at 17:43
  • Plenty of "smart" answers, you could just find $x$ and then use a calculator. – Yanko Aug 09 '17 at 17:44
  • As I said, plugging the value of x is not what I am looking for as it is quite tedious – Vindhyachal Vindhyachal Aug 09 '17 at 17:46
  • See https://math.stackexchange.com/questions/2276746/given-that-x-frac1x-sqrt3-find-x18x24 – lab bhattacharjee Aug 09 '17 at 18:13

2 Answers2

2

You might note that $x^{52}+x^{60} = x^{56}(\frac{1}{x^4}+x^4) = x^{56}.$ Similarly, $x^{48}+x^{56} = x^{52}.$ So your expression equals $x^{56}+x^{52}=x^{52}(x^4+1).$

B. Goddard
  • 1
0

$$x^{12}+1=(x^4+1)(x^8-x^4+1)=0.$$ Thus, $x^{48}+x^{60}=1-1=0$.

In another hand $x^{52}+x^{56}=x^{48}(x^4+x^8)=x^4+x^8=2x^4-1=\pm\sqrt3i$

Michael Rozenberg
  • 194,933