Let `a` is a real number satisfying `a^3+1/(a^3)=18` . Then the value of `a^4+1/(a^4)-39` is ____.

Explanation

To solve the problem, we start with the equation given:

\[ a^3 + \frac{1}{a^3} = 18 \]

### Step 1: Express \( a^3 + \frac{1}{a^3} \) in terms of \( a + \frac{1}{a} \)

We know that:

\[ a^3 + \frac{1}{a^3} = \left( a + \frac{1}{a} \right)^3 - 3 \left( a + \frac{1}{a} \right) \]

Let \( t = a + \frac{1}{a} \). Then we can rewrite the equation as:

\[ t^3 - 3t = 18 \]

### Step 2: Rearranging the equation

Rearranging gives us:

\[ t^3 - 3t - 18 = 0 \]

### Step 3: Finding the value of \( t \)

We can use trial and error to find the roots of this cubic equation. Testing \( t = 3 \):

\[ 3^3 - 3(3) - 18 = 27 - 9 - 18 = 0 \]

Thus, \( t = 3 \) is a solution.

### Step 4: Finding \( a^2 + \frac{1}{a^2} \)

Now that we have \( t = 3 \), we can find \( a^2 + \frac{1}{a^2} \) using the identity:

\[ a^2 + \frac{1}{a^2} = \left( a + \frac{1}{a} \right)^2 - 2 \]

Substituting \( t = 3 \):

\[ a^2 + \frac{1}{a^2} = 3^2 - 2 = 9 - 2 = 7 \]

### Step 5: Finding \( a^4 + \frac{1}{a^4} \)

Next, we find \( a^4 + \frac{1}{a^4} \) using the identity:

\[ a^4 + \frac{1}{a^4} = \left( a^2 + \frac{1}{a^2} \right)^2 - 2 \]

Substituting \( a^2 + \frac{1}{a^2} = 7 \):

\[ a^4 + \frac{1}{a^4} = 7^2 - 2 = 49 - 2 = 47 \]

### Step 6: Finding the final value

Now we need to find \( a^4 + \frac{1}{a^4} - 39 \):

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

Thus, the final answer is:

\[ \boxed{8} \]