If x^{2}-3x+1=0,then the value of frac{x^{6}+ | vedclass

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(C) Given equation: $x^{2}-3x+1=0$.Dividing by $x$,we get $x-3+\frac{1}{x}=0$,which implies $x+\frac{1}{x}=3$.We need to evaluate $\frac{x^{6}+x^{4}+x

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$21$

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(C) Given equation: $x^{2}-3x+1=0$.Dividing by $x$,we get $x-3+\frac{1}{x}=0$,which implies $x+\frac{1}{x}=3$.We need to evaluate $\frac{x^{6}+x^{4}+x^{2}+1}{x^{3}}$.This can be rewritten as $\frac{x^{6}}{x^{3}}+\frac{x^{4}}{x^{3}}+\frac{x^{2}}{x^{3}}+\frac{1}{x^{3}} = x^{3}+x+\frac{1}{x}+\frac{1}{x^{3}}$.Grouping the terms: $(x^{3}+\frac{1}{x^{3}}) + (x+\frac{1}{x})$.Using the identity $a^{3}+b^{3} = (a+b)^{3}-3ab(a+b)$,where $a=x$ and $b=\frac{1}{x}$:$x^{3}+\frac{1}{x^{3}} = (x+\frac{1}{x})^{3}-3(x)(\frac{1}{x})(x+\frac{1}{x}) = (3)^{3}-3(3) = 27-9 = 18$.Substituting the values: $(18) + (3) = 21$.

If $x^{2}-3x+1=0$,then the value of $\frac{x^{6}+x^{4}+x^{2}+1}{x^{3}}$ is:

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