Question

If $a = 0.01$, $b = 0.02$, $c = -0.03$ and $a + b + c = 0$, find $(a^3 + b^3 + c^3) \div (3abc)$.

• A. -1
• B. 0
• C. 1
• D. 2

Answer 1

Answer:

C. 1

Algebraic Identity

• The core principle used to solve this problem is the algebraic identity: If a + b + c = 0, then a³ + b³ + c³ = 3abc.

• This identity is derived from the factorization: a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca).

• If a + b + c = 0, the entire right-hand side of the equation becomes zero, leading to a³ + b³ + c³ - 3abc = 0, which simplifies to a³ + b³ + c³ = 3abc.

Application

• Given the values a = 0.01, b = 0.02, and c = -0.03, verify the condition: 0.01 + 0.02 + (-0.03) = 0.03 - 0.03 = 0.

• Since the condition a + b + c = 0 is satisfied, the numerator a³ + b³ + c³ is equal to 3abc.

• Substitute 3abc for a³ + b³ + c³ in the expression: (3abc) / (3abc).

• The result of the division is 1.

Key Takeaways for Competitive Exams

• Always check if a + b + c = 0 before expanding large algebraic cubes, as this specific identity frequently appears in examinations to simplify complex-looking expressions.

• This shortcut bypasses the need to calculate the actual cubes of decimal numbers, saving significant time during tests.

• Remember the broader identity: a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca), which is useful when the sum of variables is not zero.