We are given the equation:

$a + \frac{9}{a} = 3$

Our goal is to find the value of $a^3 + 27$.

Step 1: Solve for $a$

Let's first solve the given equation for $a$.

$a + \frac{9}{a} = 3$

Multiply both sides of the equation by $a$ to eliminate the denominator:

$a^2 + 9 = 3a$

Now, rearrange the terms to form a quadratic equation:

$a^2 - 3a + 9 = 0$

Step 2: Solve the quadratic equation

We can use the quadratic formula to solve for $a$. The quadratic formula is:

$a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For our equation $a^2 - 3a + 9 = 0$, the coefficients are:

• $a = 1$
• $b = -3$
• $c = 9$

Substitute these values into the quadratic formula:

$a = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(9)}}{2(1)}$ $a = \frac{3 \pm \sqrt{9 - 36}}{2}$ $a = \frac{3 \pm \sqrt{-27}}{2}$ $a = \frac{3 \pm 3i\sqrt{3}}{2}$

Thus, $a$ is a complex number, given by:

$a = \frac{3}{2} \pm \frac{3i\sqrt{3}}{2}$

Step 3: Find $a^3 + 27$

Now we need to calculate $a^3 + 27$. Instead of directly calculating the cube of this complex number, we can use the identity:

$a^3 + 27 = (a + 3)(a^2 - 3a + 9)$

We know from earlier that:

$a^2 - 3a + 9 = 0$

So:

$a^3 + 27 = (a + 3)(0) = 0$

Final Answer:

$a^3 + 27 = 0$

Let me know if you'd like any further clarification or if you have more questions!

---

Related Questions:

1. How do you solve a quadratic equation using the quadratic formula?
2. What are complex roots, and how do they appear in real-world problems?
3. How does the identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ work?
4. What are the properties of complex conjugates in equations?
5. What happens if you solve an equation with complex roots and how do you interpret the result?

Tip:

When working with complex numbers, always remember to handle the imaginary parts carefully. The presence of $\sqrt{-1}$ means the solution will involve $i$, where $i = \sqrt{-1}$.