If `(a+ 1//a) =3`, then `(a^(4) + 1//a^(4))`= ?

Options

A. 47
B. 12
C. 27
D. 49

Answer

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

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

### Step 1: Square both sides

We will square both sides of the equation to find \( a^2 + \frac{1}{a^2} \).

\[ \left(a + \frac{1}{a}\right)^2 = 3^2 \]

This expands to:

\[ a^2 + 2 \cdot a \cdot \frac{1}{a} + \frac{1}{a^2} = 9 \]

Since \( a \cdot \frac{1}{a} = 1 \), we have:

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

### Step 2: Rearranging the equation

Now, we can rearrange this equation to isolate \( a^2 + \frac{1}{a^2} \):

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

Thus, we find:

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

### Step 3: Square again to find \( a^4 + \frac{1}{a^4} \)

Next, we will square \( a^2 + \frac{1}{a^2} \) to find \( a^4 + \frac{1}{a^4} \):

\[ \left(a^2 + \frac{1}{a^2}\right)^2 = 7^2 \]

This expands to:

\[ a^4 + 2 \cdot a^2 \cdot \frac{1}{a^2} + \frac{1}{a^4} = 49 \]

Again, since \( a^2 \cdot \frac{1}{a^2} = 1 \):

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

### Step 4: Rearranging to find \( a^4 + \frac{1}{a^4} \)

Now, we can rearrange this equation to isolate \( a^4 + \frac{1}{a^4} \):

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

Thus, we find:

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

### Final Answer

Therefore, the value of \( a^4 + \frac{1}{a^4} \) is:

\[ \boxed{47} \]

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