We have the system of equations:

\[
a^3 + b^3 = 10
\]
\[
a^2 + b^2 = 7
\]

Let's solve for \( a + b \).

### Step 1: Express the identity
The identity for the sum of cubes is:
\[
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
\]
We know from the second equation that:
\[
a^2 + b^2 = 7
\]
Thus, we can write:
\[
a^2 + b^2 = (a + b)^2 - 2ab = 7
\]
Let's set \( s = a + b \) and \( p = ab \). Then:
\[
s^2 - 2p = 7 \quad \text{(Equation 1)}
\]
And for the sum of cubes:
\[
a^3 + b^3 = s(s^2 - 3p) = 10 \quad \text{(Equation 2)}
\]

### Step 2: Solve for \( p \) using Equation 1
\[
p = \frac{s^2 - 7}{2}
\]

### Step 3: Substitute \( p \) into Equation 2
Substitute \( p \) into the second equation:
\[
s(s^2 - 3p) = 10
\]
\[
s\left(s^2 - 3\left(\frac{s^2 - 7}{2}\right)\right) = 10
\]
Multiply through to eliminate the fraction:
\[
s\left(s^2 - \frac{3}{2}(s^2 - 7)\right) = 10
\]
\[
s\left(s^2 - \frac{3s^2 - 21}{2}\right) = 10
\]
\[
s\left(\frac{2s^2 - 3s^2 + 21}{2}\right) = 10
\]
\[
s\left(\frac{-s^2 + 21}{2}\right) = 10
\]
\[
s(-s^2 + 21) = 20
\]
\[
-s^3 + 21s = 20
\]

### Step 4: Solve the cubic equation
\[
s^3 - 21s + 20 = 0
\]

By trial, synthetic division, or factoring, the solution to this cubic equation is \( s = 1 \). 

### Final Answer
\[
a + b = 1
\]

Would you like more details, or do you have any questions?

---

Here are five related questions for practice:
1. If \( a + b = 1 \), what are the possible values of \( ab \)?
2. How would the solution change if \( a^3 + b^3 = 15 \)?
3. Can you find the individual values of \( a \) and \( b \)?
4. What is the relationship between \( a \) and \( b \) given the solution \( a + b = 1 \)?
5. How does the discriminant of a cubic equation help in finding the roots?

**Tip:** For solving cubic equations, trying simple integers as possible roots can often simplify the process significantly.

Learn how to solve the system of equations involving cubes and quadratics: a^3 + b^3 = 10, a^2 + b^2 = 7. This problem explores the identities of cube sums and quadratic equations, suitable for advanced high school students.

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