https://file.aiquickdraw.com/mathbot/file/1727421641791a8zgv7xh.jpg

The system of equations provided in the image is:

\[
\begin{cases}
a^3 + b^3 = 133 \quad \text{(1)} \\
a^2 - ab + b^2 = 19 \quad \text{(2)}
\end{cases}
\]

Let's solve this system step by step.

### Step 1: Factorize the first equation
We can factor \( a^3 + b^3 \) using the identity:
\[
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
\]
This transforms equation (1) into:
\[
(a + b)(a^2 - ab + b^2) = 133
\]

But from equation (2), we already know that:
\[
a^2 - ab + b^2 = 19
\]

Substitute this value into the factorized equation:
\[
(a + b) \times 19 = 133
\]
Solving for \( a + b \):
\[
a + b = \frac{133}{19} = 7
\]

### Step 2: Solve the system
Now, we have the following simplified system:
\[
\begin{cases}
a + b = 7 \\
a^2 - ab + b^2 = 19
\end{cases}
\]

We can substitute \( b = 7 - a \) into the second equation.

### Step 3: Substitute into the second equation
Substitute \( b = 7 - a \) into equation (2):
\[
a^2 - a(7 - a) + (7 - a)^2 = 19
\]
Simplify this expression:
\[
a^2 - 7a + a^2 + (7 - a)^2 = 19
\]
Expanding \( (7 - a)^2 \):
\[
(7 - a)^2 = 49 - 14a + a^2
\]
So, we get:
\[
a^2 - 7a + a^2 + 49 - 14a + a^2 = 19
\]
Combine like terms:
\[
3a^2 - 21a + 49 = 19
\]
Subtract 19 from both sides:
\[
3a^2 - 21a + 30 = 0
\]
Divide through by 3:
\[
a^2 - 7a + 10 = 0
\]

### Step 4: Solve the quadratic equation
The quadratic equation is:
\[
a^2 - 7a + 10 = 0
\]
We can solve this using the quadratic formula:
\[
a = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(10)}}{2(1)}
\]
Simplifying:
\[
a = \frac{7 \pm \sqrt{49 - 40}}{2}
\]
\[
a = \frac{7 \pm \sqrt{9}}{2}
\]
\[
a = \frac{7 \pm 3}{2}
\]

Thus, we have two solutions for \( a \):
\[
a = \frac{7 + 3}{2} = 5 \quad \text{or} \quad a = \frac{7 - 3}{2} = 2
\]

### Step 5: Find corresponding values for \( b \)
- If \( a = 5 \), then \( b = 7 - 5 = 2 \).
- If \( a = 2 \), then \( b = 7 - 2 = 5 \).

Thus, the solutions are:
\[
(a, b) = (5, 2) \quad \text{or} \quad (a, b) = (2, 5)
\]

Would you like more details on any step? Here are some related questions to deepen understanding:

1. How does the factorization of sums of cubes help simplify equations?
2. Why is \( a^2 - ab + b^2 = 19 \) helpful for substitution in this problem?
3. Can you always use the quadratic formula to solve second-degree polynomials?
4. How do different identities (like the sum of cubes) help in solving algebraic equations?
5. What are the geometric interpretations of equations like \( a^3 + b^3 = 133 \)?

**Tip:** Always look for common algebraic identities that can simplify complex expressions, such as factoring or using formulas like the sum of cubes.

Solve the system of equations: a^3 + b^3 = 133 and a^2 - ab + b^2 = 19

Learn how to solve the system of equations a^3 + b^3 = 133 and a^2 - ab + b^2 = 19 using factoring and the quadratic formula. Detailed steps include the sum of cubes identity and solving a quadratic equation, suitable for high school algebra students in Grades 10-12.

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