Question

If $x^4+\frac{1}{x^4}=119,\text{find the value of}{x}^3-\frac{1}{x^3}$.

Answer 1

$x^4+\frac{1}{x^4}=119$

Adding 2 to both sides,

$x^4+\frac{1}{x^4}+2=119+2\Rightarrow (x^2{)}^2+{\left(\frac{1}{x^2}\right)}^2+2=121{(x^2+\frac{1}{x^2})}^2=(11{)}^2\Rightarrow x^2+\frac{1}{x^2}=11x^2+\frac{1}{x^2}=11\text{Subtracting 2 from both sides,}{x}^2+\frac{1}{x^2}-2=11-2=9\Rightarrow {(x-\frac{1}{x})}^2=(3{)}^2\Rightarrow x-\frac{1}{x}=3\Rightarrow x-\frac{1}{x}=3\text{Cubing both sides}{(x-\frac{1}{x})}^3=(3{)}^3{x}^3-\frac{1}{x^3}-3(x-\frac{1}{x})=27\Rightarrow x^3-\frac{1}{x^3}-3\times 3=27\Rightarrow x^3-\frac{1}{x^3}-9=27\therefore x^3-\frac{1}{x^3}=27+9=36$