How can I calculate the value of $a^4+ \frac {1}{a^4}$?

Question

given that $a$ is real positive number such that $a^3+ \frac {1}{a^3}=18$ How can I calculate the value of $a^4+ \frac {1}{a^4}$ ?

What is the context of this problem? Have you tried to solve it? If so, what difficulty did you actually encounter? Supplying these details will assist a willing Reader to provide an Answer suitable to your level of effort and understanding. — hardmath, Feb 25 '15 at 21:49

score 3 · Answer 1 · edited Feb 25 '15 at 20:58

3

Let $x \stackrel{\rm def}{=}a + \frac{1}{a}$. First, show that $x^3 = 3x + 18$; which implies $x = 3$. From there, we get (by computing $x^2$ then $x^4$) $$\frac{1}{a^2} + a^2 = 7$$ and $$\frac{1}{a^4} + a^4 = 47.$$

edited Feb 25 '15 at 20:58

Clement C.

67,323

answered Feb 25 '15 at 20:10

Lozenges

1,742

How did you see x=3 without graphing the equation? This is where I got stuck. – Joe Slater Feb 25 '15 at 20:14
Nice solution. See this page for how to write math on this site. – Winther Feb 25 '15 at 20:19

score 2 · Answer 2 · answered Feb 25 '15 at 19:49

2

Hint: find the value of $a>0$. To do so, it suffices to find the value of $a^3$: do you know how to solve $A+\frac{1}{A}=\alpha$ for $A>0$?

answered Feb 25 '15 at 19:49

Clement C.

67,323

I really dont know how but if u told me I will try it in this exercice :D – u100899 Feb 25 '15 at 19:54
Multiply both sides by $A$ and rearrange the terms: you only need to solve $A^2-\alpha A + 1=0$ for $A$. – Clement C. Feb 25 '15 at 19:57
This solution will require a calculator (to solve using quadratic formula). Is there no way to solve this without calculator? I suspect there must be since the solution is coming as an integer - 47 . – Joe Slater Feb 25 '15 at 19:58
You can write the expression and try to simplify it without a calculator. – Clement C. Feb 25 '15 at 20:01
I tried. I am not able to. – Joe Slater Feb 25 '15 at 20:03
@Joe: There may very well be a simpler solution (not involving the computation of $a$) — I do believe there is. For reference, the answer is 47. – Clement C. Feb 25 '15 at 20:05

barak manos · Answer 3 · 2015-02-25T20:01:02.913

2

Denote $t=a^3$, then:

$a^3+\frac{1}{a^3}=18\implies$

$t+\frac{1}{t}=18\implies$

$t^2-18t+1=0\implies$

$t=\frac{18\pm\sqrt{324-4}}{2}\implies$

$a^3=\frac{18\pm\sqrt{324-4}}{2}\implies$

$a=\sqrt[3]{\frac{18\pm\sqrt{324-4}}{2}}\implies$

$a^4+\frac{1}{a^4}=\sqrt[3]{\frac{18\pm\sqrt{324-4}}{2}}^4+\frac{1}{\sqrt[3]{\frac{18\pm\sqrt{324-4}}{2}}^4}=47$

edited Feb 25 '15 at 20:01

answered Feb 25 '15 at 19:55

barak manos

43,109

score 2 · Answer 4 · answered Feb 25 '15 at 20:25

$a^4+\dfrac1{a^4}=\big(a^2\big)^2+\bigg(\dfrac1{a^2}\bigg)^2=\bigg(a^2+\dfrac1{a^2}\bigg)^2-2.\quad$ Similarly, $a^2+\dfrac1{a^2}=\bigg(a+\dfrac1a\bigg)^2-2$.

At the same time, $a^3+\dfrac1{a^3}=\bigg(a+\dfrac1a\bigg)\bigg(a^2+\dfrac1{a^2}-1\bigg).~$ Letting $S_n=a^n+\dfrac1{a^n}~,~$ we have

$~S_2=S_1^2-2,\quad~S_3=S_1~\Big(S_2-1\Big)=S_1~\Big(S_1^2-3\Big)=18,\quad~S_4=S_2^2-2.~$ It is trivial to

notice that $S_1=3$ is a solution for $S_3=18,~$ implying $S_2=7~$ and $~S_4=47$.

How can I calculate the value of $a^4+ \frac {1}{a^4}$?

4 Answers4

Linked