Math Series #2

For any number $n$ , let ${S_n}=x^n+\dfrac{1}{x^n}$ . So, we have, $\begin{aligned} S_1&=3 \\ \\ S_2&=\left(x+\frac{1}{x}\right)^2-2\cdot x\cdot \frac{1}{x} \\ &=S_1^2-2 \\ &=7 \\ \\ S_3&=\left(x+\frac{1}{x}\right)^3-3\cdot x\cdot \frac{1}{x}\cdot \left(x+ \frac{1}{x} \right) \\ &=S_1^3-3S_1 \\ &=18 \\ \\ S_4&=\left(x^2+\frac{1}{x^2}\right)^2-2\cdot x^2\cdot \frac{1}{x^2} \\ &=S_2^2-2 \\ &=47 \\ \\ x^4+x^3+\frac{1}{x^4}+\frac{1}{x^3}&=\left(x^4+\frac{1}{x^4}\right)+\left(x^3+\frac{1}{x^3}\right) \\ &=S_4+S_3 \\ &=47+18 \\ &=\boxed{65} \end{aligned}$

$(x + \frac {1}{x})^2 = x^2 + \frac {1}{x^2} +(2 \times x \times \frac {1}{x}) = x^2 + \frac {1}{x^2} + 2 = 3^2 = 9 \rightarrow x^2 + \frac {1}{x^2} = 9 - 2 = 7 \rightarrow (x^2 + \frac {1}{x^2})^2 = x^4 + \frac {1}{x^4} + (2 \times x^2 + \frac {1}{x^2})= x^4 + \frac {1}{x^4} + 2 = 7^2 = 49 \rightarrow x^4 + \frac {1}{x^4} = 49 - 2 = 47 \rightarrow (x^2 + \frac {1}{x^2}) \times (x + \frac {1}{x}) = x^3 + \frac {1}{x^3} + x + \frac {1}{x} = x^3 + \frac {1}{x^3} + 3 = 7 \times 3 = 21 \rightarrow x^3 + \frac {1}{x^3} = 21 - 3 = 18 \rightarrow x^4 + x^3 + \frac {1}{x^4} + \frac {1}{x^3} = 47 + 18 = \boxed{65}$