Question

If $\sqrt[3]{x}+\frac{36}{\sqrt[3]{x}}=13$, then how many integer values of $x$ satisfy the equation?

Answer 1

Answer: 2744

$\sqrt[3]{x}+\frac{18}{\sqrt[3]{x}}=9$
$\sqrt[3]{x}$ should be a factor of 18 .
Hence, $\sqrt[3]{x}$ is either $1,2,3,4,6,9,12$, or 18 .
Since $\sqrt[3]{x}+$ some number is 9 .
$\sqrt[3]{x}$ should be less than 9 and greater than 3.
Let $\sqrt[3]{x}=4$
$\sqrt[3]{x}+\frac{18}{\sqrt[3]{x}}=13$
$\text{ LHS }=4+\frac{18}{4}$
$\ne 4+\frac{9}{2}=\frac{17}{2}\ne \text{ RHS }$
If $\sqrt[3]{x}=6$, then $\sqrt[3]{x}+\frac{18}{\sqrt[3]{x}}=6+\frac{18}{6}$
$=6+3=9=\text{ RHS }$
$\therefore \sqrt[3]{x}=6$
Only one integer value of $x$ satisfy the equation.