Question

If $x+\frac{1}{x}=3$, calculate $x^2+\frac{1}{x^2},x^3+\frac{1}{x^3}and{x}^4+\frac{1}{x^4}$.

Answer 1

(i) $x+\frac{1}{x}=3$

Squaring both sides,

${(x+\frac{1}{x})}^2=(3{)}^2\Rightarrow x^2+\frac{1}{x^2}+2=9\Rightarrow x^2+\frac{1}{x^2}=9-2=7\therefore x^2+\frac{1}{x^2}=7(ii)x+\frac{1}{x}=3\text{Cubing both sides}{(x+\frac{1}{x})}^3=(3{)}^3{x}^3+\frac{1}{x^3}+3(x+\frac{1}{x})=27x^3+\frac{1}{x^3}+3\times 3=27\Rightarrow x^3+\frac{1}{x^3}+9=27\Rightarrow x^3+\frac{1}{x^3}=27-9=18\therefore x^3+\frac{1}{x^3}=18\left(iii\right)x^2+\frac{1}{x^2}=7\text{Squaring both sides},{(x^2+\frac{1}{x^2})}^2=(7{)}^2\Rightarrow x^4+\frac{1}{x^4}+2=49\Rightarrow x^4+\frac{1}{x^4}=49-2=47\therefore x^4+\frac{1}{x^4}=47$