Title: Solving the Equation –3y(x² + 4xy − 5y²) = −3x²y − 12xy² + 15y³ Step-by-Step

Introduction
Understanding and simplifying algebraic equations is essential for students, researchers, and mathematics enthusiasts alike. One such equation—–3y(x² + 4xy − 5y²) = −3x²y − 12xy² + 15y³—can seem complex at first glance, but with a clear step-by-step breakdown, solving it becomes manageable. This article walks you through the full process of simplifying, expanding, and analyzing the equation to solve it clearly, alongside a practical example and key takeaways.

Understanding the Context

Understanding the Equation
We begin with:
–3y(x² + 4xy − 5y²) = −3x²y − 12xy² + 15y³
This is a bilinear equation in two variables x and y. The left-hand side (LHS) is a product involving y and a quadratic trinomial in x and y, while the right-hand side (RHS) is a cubic polynomial in x and y, including a pure cubic term in y³.

Rewriting the equation in standard form:

→ Move every term to the left-hand side:
–3y(x² + 4xy − 5y²) + 3x²y + 12xy² − 15y³ = 0

-3y(x^2 + 4xy - 5y^2) = -3x^2y - 12xy^2 + 15y^3

Key Insights

Step 1: Expand the Left Side
Distribute –3y across the parentheses:
–3y·x² − 3y·4xy + 3y·5y² + 3x²y + 12xy² − 15y³
→ –3x²y − 12xy² + 15y³ + 3x²y + 12xy² − 15y³

Step 2: Combine Like Terms
Group similar terms:

  • x²y: (–3x²y + 3x²y) = 0
  • xy²: (–12xy² + 12xy²) = 0
  • y³: (15y³ – 15y³) = 0

Thus, the entire expression simplifies to:
0 = 0

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Final Thoughts

Step 3: Interpret the Result
Since both sides reduce to zero, the equation –3y(x² + 4xy − 5y²) = −3x²y − 12xy² + 15y³ is identically true for all real numbers x and y. In other words, the equation represents a trivial identity — it holds universally.

This means every ordered pair (x, y) satisfies the equation, and thus the expression is an algebraic identity.

Example Substitution (for clarity)
Choose x = 1, y = 2:

  • LHS: –3(2)(1² + 4(1)(2) – 5(2)²) = –6(1 + 8 – 20) = –6(–11) = 66
  • RHS: –3(1)²(2) – 12(1)(2)² + 15(2)³ = –6 – 48 + 120 = 66
    ✅ Both sides equal 66 — confirming the identity.

Why This Matters (Mathematical Insight)
Recognizing identities like this streamlines problem-solving. It avoids unnecessary algebraic steps and clarifies relationships between variables. Identities simplify calculus operations, parametric modeling, and error-checking in applied mathematics.

Key Takeaways

  • Expand and simplify expressions carefully.
  • Combine like terms accurately.
  • Recognizing identical forms (like 0 = 0) indicates an identity.
  • Substituting sample values verifies correctness.
  • This equation is universally true for all real x, y.