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Question

$x^2 + \frac{1}{x^2}=34$ and $x$ is a natural number. Find the value of $x^3 + \frac{1}{x^3}$ and choose the correct answer from the following options:

  1. 198
  2. 216
  3. 200
  4. 186

What I have did yet

I tried to find the value of $x + \frac{1}{x}$. Here are my steps to do so: $$x^2 + \frac{1}{x^2} = 34$$ $$\text{Since}, (x+\frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}$$ $$\Rightarrow (x+\frac{1}{x})^2-2=34$$ $$\Rightarrow (x+\frac{1}{x})^2=34+2 = 36$$ $$\Rightarrow x+\frac{1}{x}=\sqrt{36}=6$$

I have calculated the value of $x+\frac{1}{x}$ is $6$. I do not know what to do next.

Starship - On Strike
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Abhigyan Kumar
  • 159
  • What happens when you cube $x+\frac 1x$? – Mark Bennet Jan 14 '21 at 11:06
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    Note that $x$ cannot be a "natural number", and you are missing squaring/taking the square root in some of your calculations. You are, though on the right track, but you need to take a little more care. – Mark Bennet Jan 14 '21 at 11:09
  • Sorry, but according to the question $x$ will be a natural number. – Abhigyan Kumar Jan 14 '21 at 11:12
  • If $x$ is a natural number other than $1$ then $x^2+\frac 1{x^2}$ cannot be an integer. And $x\neq 1$ is easy to check. It doesn't actually matter very much here. – Mark Bennet Jan 14 '21 at 11:16

3 Answers3

5

Hint $(x+\frac{1}{x})^3=x^3+\frac{1}{x^3}+3(x+\frac{1}{x})$ also $(x+\frac{1}{x})=6$ not $36$ you seem to have typed an incorrect calculation.

Archis Welankar
  • 15,910
1

You wrote

$$(x+\frac{1}{x}) = x^2 + 2 + \frac{1}{x^2}.$$

But it should read

$$(x+\frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}.$$

Then we obtain $x+\frac{1}{x}=6.$

From $x^2 + \frac{1}{x^2}=34$ we obtain

$(1) \quad x^3+\frac{1}{x}=34 x$

and

$(2) \quad x+\frac{1}{x^3}=\frac{34}{x}.$

If we add (1) and (2) we get

$$x^3+\frac{1}{x^3}=33(x+\frac{1}{x}).$$

Ans so

$$x^3+\frac{1}{x^3}=6 \cdot 33=198.$$

Remark : if $x^2 + \frac{1}{x^2}=34$, then $x$ can not be a natural number !

Fred
  • 77,394
1

Let have a look at this: finding $x^6+y^6$ given $x+y$ and $xy$

$$U_n=x^n+\frac 1{x^n}$$

$s=x+\frac 1x=U_1$ and $p=x\times\frac 1x=1$

also we have $\begin{cases}U_2=34\\U_0=x^0+\frac 1{x^0}=1+1=2\end{cases}$

We have the relation $$U_{n+1}=U_1U_n-U_{n-1}$$

$U_2=U_1^2-U_0\implies U_1^2=34+2=36\implies U_1=\pm 6$

And $U_3=U_1U_2-U_1=\pm 6(34-1)=\pm 198\quad$ (among proposed answers the positive one fits).

Note: $x$ cannot be integer, since $x+\frac 1x=\pm 6\iff x=\pm(3\pm 2\sqrt{2})$

This method can be generalized to most $x^n+y^n$ types of questions.

zwim
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