Question

Let $a$ is real number satisfying $a^3+\frac{1}{a^3}=18$. Then the value of $a^4+\frac{1}{a^4}-39$ is _______.

Answer 1

Let $(a+\frac{1}{a})=t$

$\Rightarrow a^3+\frac{1}{a^3}=18$

${(a+\frac{1}{a})}^3-3(a+\frac{1}{a})=18$ ..................... $(a+b{)}^3=a^3+3ab(a+b)+b^3$

$t^3-3t-18=0$ ......(1)

$t=3$ satisfies (1)

hence factorizing (1)

$(t-3)(t^2+3t+6)=0$

$t=3$ only is the solution

$\therefore a+\frac{1}{a}=3\Rightarrow a^2+\frac{1}{a^2}=7$
$\Rightarrow a^4+\frac{1}{a^4}=47$

$a^4+\frac{1}{a^4}-39=47-39=8$