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Given $x+\frac{1}{x}=2 \tag{1}$
$x^3+\frac{1}{x^4}=1 \tag{2}$

Find the value of $$x^4+\frac{1}{x^3}$$

If equation $\bf{2}$ is multiplied by $\bf x$ we get $x^4+\frac{1}{x^3}=x$ now, we just need to find the value of $x$ which from $\bf{1}$ is equal to $1$. But the answer to this question is $\bf3$. Which can be get by solving differently. Why my method is wrong is it because I am introducing new root into the equation.

Another approach Given, $(x + 1/x) = 2$

Squaring both sides,

$$x^2 + 1/x^2 + 2 = 4$$

$$ x^2 + 1/x^2 = 2 $$

Let,

$A = x^3 + 1/x^4$

$B = x^4 + 1/x^3$

Now, add A and B,

$A + B = x^3 + 1/x^4 + x^4 + 1/x^3$

$A + B = x^3 + 1/x^3 + x^4 + 1/x^4$

$A + B = (x + 1/x)(x^2 + 1/x^2 – 1) + (x^2)2 + (1/x^2)2 + 2 – 2$

$A + B = 2(2 – 1) + (x^2 + 1/x^2)2 – 2 $

$A + B = 2 + (2)2 – 2$

$ A + B = 4$

Given that $A = 1$

Then, $B = 4 – 1 = 3$

$ x^4 + 1/x^3 = 3$

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    $\begingroup$ If $x+\frac1x=2$ then $x=1$. Are you sure you have the question right? $\endgroup$ – Lord Shark the Unknown Jul 22 at 17:10
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The system is inconsistent because the first equation gives $x=1$ which does not satisfy the second equation.

Thus the system does not have a solution.

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  • $\begingroup$ Oh thank you for figuring it out I missed it. I will delete this question because its wrong $\endgroup$ – Nebo Alex Jul 22 at 17:18
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    $\begingroup$ It is a good question. Keep it alive $\endgroup$ – Mohammad Riazi-Kermani Jul 22 at 17:22
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If $x=3$ and $x$ is solution to $(1)$, we have $$ x + \frac{1}{x} = 2 \iff 3 + \frac 1 3 = 2 $$ which is obviously wrong. There must be some mistake in the solution.

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