ambiguity in solving algebraic equations. 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$

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. 
