If `x=a+b i` is a complex number such that `x^2=3+4i and x^3=2+1i ,w h e r e i=sqrt(-1),t h e n(a+b)` equal to ______.

Explanation

To solve the problem step by step, we need to find the values of \( a \) and \( b \) in the complex number \( x = a + bi \) given the conditions \( x^2 = 3 + 4i \) and \( x^3 = 2 + i \).

### Step 1: Express \( x \) in terms of \( x^2 \) and \( x^3 \)
We can express \( x \) as:
\[ x = \frac{x^3}{x^2} \]
Given that \( x^3 = 2 + i \) and \( x^2 = 3 + 4i \), we substitute these values:
\[ x = \frac{2 + i}{3 + 4i} \]

**Hint:** To simplify the division of complex numbers, multiply the numerator and denominator by the conjugate of the denominator.

### Step 2: Multiply by the conjugate of the denominator
The conjugate of \( 3 + 4i \) is \( 3 - 4i \). We multiply both the numerator and denominator by this conjugate:
\[ x = \frac{(2 + i)(3 - 4i)}{(3 + 4i)(3 - 4i)} \]

**Hint:** Remember that the product of a complex number and its conjugate gives a real number.

### Step 3: Calculate the denominator
Calculating the denominator:
\[ (3 + 4i)(3 - 4i) = 3^2 - (4i)^2 = 9 - (-16) = 9 + 16 = 25 \]

**Hint:** Use the formula \( a^2 - b^2 \) for the product of a complex number and its conjugate.

### Step 4: Calculate the numerator
Now calculate the numerator:
\[ (2 + i)(3 - 4i) = 2 \cdot 3 + 2 \cdot (-4i) + i \cdot 3 + i \cdot (-4i) = 6 - 8i + 3i - 4(-1) \]
Simplifying this gives:
\[ 6 - 8i + 3i + 4 = 10 - 5i \]

**Hint:** Combine like terms carefully, remembering that \( i^2 = -1 \).

### Step 5: Combine results
Now we can combine the results:
\[ x = \frac{10 - 5i}{25} = \frac{10}{25} - \frac{5}{25}i = \frac{2}{5} - \frac{1}{5}i \]

**Hint:** Break down the fraction into real and imaginary parts.

### Step 6: Identify \( a \) and \( b \)
From \( x = a + bi \), we can identify:
\[ a = \frac{2}{5}, \quad b = -\frac{1}{5} \]

### Step 7: Calculate \( a + b \)
Now, we find \( a + b \):
\[ a + b = \frac{2}{5} - \frac{1}{5} = \frac{1}{5} \]

### Final Answer
Thus, the value of \( a + b \) is:
\[ \boxed{\frac{1}{5}} \]