if $$ \sqrt[3]{x}+\frac{1}{\sqrt[3]{x}}=9 $$, then $$ x+\frac{1}{x}= $$ ?

Question

UpStudy AI · Accepted Answer

Let $$ t = \sqrt[3]{x} $$. Then the given equation becomes

$$
t + \frac{1}{t} = 9.
$$

We want to find

$$
x + \frac{1}{x} = t^3 + \frac{1}{t^3}.
$$

Recall the identity

$$
\left(t + \frac{1}{t}\right)^3 = t^3 + \frac{1}{t^3} + 3\left(t + \frac{1}{t}\right).
$$

Substitute $$ t + \frac{1}{t} = 9 $$:

$$
9^3 = t^3 + \frac{1}{t^3} + 3 \times 9.
$$

This simplifies to:

$$
729 = t^3 + \frac{1}{t^3} + 27.
$$

Subtract 27 from both sides:

$$
t^3 + \frac{1}{t^3} = 729 - 27 = 702.
$$

Thus, the value of $$ x + \frac{1}{x} $$ is

$$
\boxed{702}.
$$
$$ x + \frac{1}{x} = 702 $$

if \( \sqrt[3]{x}+\frac{1}{\sqrt[3]{x}}=9 \), then \( x+\frac{1}{x}= \) ?