If x = 2 + √3 , then the value of $$\frac{x^6 + x^4 + x^2 + 1}{x^3}$$is
We have x=$$\ 2+\sqrt{\ 3}$$
$$\ \frac{1}{x}=\frac{1}{2+\sqrt{\ 3}}=2-\sqrt{\ 3}$$
We have to find $$x^3+\frac{1}{x^3}+x+\frac{1}{x}$$
Now x+1/x =4
Also $$x^3+\frac{1}{x^3}=\ 4^3-3\left(x+\frac{1}{x}\right)=52$$
so we get x^3+1/x^3+x+1/x =56
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