Question

Se x+$\frac{1}{x}$ =12, calcule x^2 +$\frac{1}{x^2}$   e  x³+$\frac{1}{x^3}$

Answer 1

$\left(x+\frac{1}{x}\right)^2=x^2+2+\frac{1}{x^2}=144\\ \\ x^2+\frac{1}{x^2}=144-2\\ \\ \boxed{x^2+\frac{1}{x^2}=142}$

$\left(x+\frac{1}{x}\right)^3=\left(x^3+\frac{1}{x^3}\right)+3.\left(x+\frac{1}{x}\right)\\ \\ 1728=\left(x^3+\frac{1}{x^3}\right)+3.12\\ \\ \left(x^3+\frac{1}{x^3}\right)=1728-36\\ \\ \boxed{\left(x^3+\frac{1}{x^3}\right)=1692}$