Given $$ x^2-3x+1=0 $$ , find the value of the expression $$ x^2+\dfrac{1}{x^2} $$ , $$ x^3+\dfrac{1}{x^3} $$ and $$ x^4+\dfrac{1}{x^4} $$

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Uniteasy Admin · Accepted Answer

$$ \because x^2-3x+1=0 $$
$$ 	herefore x+\dfrac{1}{x} =3 $$n
Square both sides, we get,
$$ x^2+\dfrac{1}{x^2} +2 = 9 $$
$$ 	herefore  x^2+\dfrac{1}{x^2} = 7 $$n
Multiply equation (1) with (2),
$$ x^3+\dfrac{1}{x}+x+\dfrac{1}{x^3} = 21 $$
Plug in (1), we get
$$ x^3+\dfrac{1}{x^3} = 18 $$n
Square both sides of equation (2), we get
$$ x^4+\dfrac{1}{x^4}+2 = 49 $$
$$ 	herefore  x^4+\dfrac{1}{x^4} = 47 $$n
Now we have come to the end of our conclusion.
Given the condistion $$ x+\dfrac{1}{x} =3 $$ ,
$$ \begin{cases} x^2+\dfrac{1}{x^2} = 7 &  \ 
\
x^3+\dfrac{1}{x^3} = 18 & \ 
\
 x^4+\dfrac{1}{x^4} = 47
\end{cases} $$

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