%5E3-formula-expansion.jpeg "(a-b-c)^3 Formula Expansion")
Understanding the Expansion of (a-b-c)³
The expansion of the expression (a-b-c)³ might seem daunting at first glance, but it's a straightforward application of the binomial theorem and a little bit of pattern recognition. This article will break down the process step-by-step, providing a clear understanding of the formula and its application.
The Binomial Theorem
The binomial theorem is a powerful tool for expanding expressions of the form (x+y)ⁿ. It states that:
(x + y)ⁿ = ∑(n choose k) * x^(n-k) * y^k
where:
- n is a non-negative integer (the power)
- k is an integer ranging from 0 to n
- (n choose k) is the binomial coefficient, calculated as n! / (k! * (n-k)!)
Expanding (a-b-c)³
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Reframing the Expression: We can rewrite (a-b-c)³ as [(a-b) - c]³. This allows us to treat (a-b) as a single term.
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Applying the Binomial Theorem: We'll use the binomial theorem twice:
- First, for expanding [(a-b) - c]³, treating (a-b) as x and -c as y.
- Then, we'll use it again to expand (a-b)².
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Step-by-Step Expansion:
- [(a-b) - c]³ = (a-b)³ + 3(a-b)²(-c) + 3(a-b)(-c)² + (-c)³
- = (a-b)³ - 3c(a-b)² + 3c²(a-b) - c³
- = [a³ - 3a²b + 3ab² - b³] - 3c[a² - 2ab + b²] + 3c²(a-b) - c³
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Simplifying:
- = a³ - 3a²b + 3ab² - b³ - 3a²c + 6abc - 3b²c + 3ac² - 3bc² - c³
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Final Result:
- (a-b-c)³ = a³ - 3a²b + 3ab² - b³ - 3a²c + 6abc - 3b²c + 3ac² - 3bc² - c³
Key Observations
- Alternating Signs: Notice that the terms in the expanded form alternate in sign.
- Coefficients: The coefficients of the terms follow the pattern of Pascal's triangle (1, 3, 3, 1) due to the binomial theorem.
- Powers: The powers of a, b, and c decrease as the terms move from left to right.
Applications
The expansion of (a-b-c)³ is useful in various mathematical contexts, including:
- Algebraic Simplification: It helps simplify complex algebraic expressions.
- Polynomial Expansion: It plays a role in expanding polynomials with multiple variables.
- Calculus: It can be used for finding derivatives and integrals of complex functions.
By understanding the formula and its derivation, you can confidently expand similar expressions and apply it in various mathematical scenarios.