Math Problem Statement

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

Solution

The system of equations provided in the image is:

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.

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Math Problem Analysis

Mathematical Concepts

Algebra
System of Equations
Factoring

Formulas

Sum of cubes: a^3 + b^3 = (a + b)(a^2 - ab + b^2)
Quadratic formula: a = (-b ± √(b² - 4ac)) / 2a

Theorems

Sum of cubes
Quadratic formula

Suitable Grade Level

Grades 10-12

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