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$\\Solution:\\\\ Given: x+\frac1x=5\\\\ \text{Squaring on both sides}\\\\ \Rightarrow (x+\frac1x)^2=5^2\\\\ \Rightarrow x^2+2\times x\times\frac1x+\left (\frac{1}{x} \right )^2=25\ \ \ \ \ \ \ \ [\because (a+b)^2=a^2+b^2+2ab]\\\\ \Rightarrow x^2+2+\frac{1}{x^2}=25\\\\ \Rightarrow x^2+\frac{1}{x^2}=25-2\\\\ \therefore \left (x^2+\frac{1}{x^2} \right )=23\\\\ \text{Squaring on both sides}\\\\ \Rightarrow \left (x^2+\frac{1}{x^2} \right )^2=23^2\\ \Rightarrow x^4+2\times x^2\times\frac1{x^2}+\left (\frac{1}{x^2} \right )^2=529\ \ \ \ \ \ \ \ [\because (a+b)^2=a^2+b^2+2ab]\\\\ \Rightarrow x^4+2+\frac{1}{x^4}=529\\\\ \Rightarrow x^4+\frac{1}{x^4}=529-2\\\\ \therefore \left (x^4+\frac{1}{x^4} \right )=527$

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$\\Solution:\\\\ Given:\\\\ x^4+x^2+1\\\\ Adding\ and\ subtracting\ x^2\ to\ the\ given\ equation\\\\ =x^4+x^2+1+x^2-x^2\\\\ =x^4+2x^2+1-x^2\\\\ =(x^2+1)^2-x^2\ \ \ \ \ \ \ \ \ \ [\because a^2+2ab+b^2=(a+b)^2]\\\\ =(x^2+1+x)((x^2+1-x))\ \ \ \ \ \ \ [\because a^2-b^2=(a+b)(a-b)]$

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