%5E3-expand.jpeg "(x+3)^3 Expand")
Expanding (x+3)³
The expression (x+3)³ represents the cube of the binomial (x+3). Expanding this means writing it out as a sum of terms. There are two main ways to do this:
1. Using the Binomial Theorem
The binomial theorem provides a general formula for expanding any binomial raised to a power. The formula states:
(a+b)ⁿ = ⁿC₀aⁿb⁰ + ⁿC₁aⁿ⁻¹b¹ + ⁿC₂aⁿ⁻²b² + ... + ⁿCₙa⁰bⁿ
where ⁿCᵣ represents the binomial coefficient, calculated as n! / (r! * (n-r)!).
Applying this to (x+3)³, we get:
(x+3)³ = ³C₀x³3⁰ + ³C₁x²3¹ + ³C₂x¹3² + ³C₃x⁰3³
Calculating the binomial coefficients:
- ³C₀ = 3! / (0! * 3!) = 1
- ³C₁ = 3! / (1! * 2!) = 3
- ³C₂ = 3! / (2! * 1!) = 3
- ³C₃ = 3! / (3! * 0!) = 1
Substituting these values back into the equation:
(x+3)³ = 1x³1 + 3x²3 + 3x9 + 1127
Finally, simplifying the expression:
(x+3)³ = x³ + 9x² + 27x + 27
2. Expanding by Repeated Multiplication
We can also expand (x+3)³ by repeatedly multiplying the binomial by itself:
(x+3)³ = (x+3)(x+3)(x+3)
First, we multiply the first two factors:
(x+3)(x+3) = x² + 3x + 3x + 9 = x² + 6x + 9
Then, we multiply this result by the remaining (x+3):
(x² + 6x + 9)(x+3) = x³ + 3x² + 6x² + 18x + 9x + 27
Finally, combining like terms:
(x+3)³ = x³ + 9x² + 27x + 27
Conclusion
Both methods lead to the same result: (x+3)³ = x³ + 9x² + 27x + 27
The binomial theorem provides a more general approach for expanding any binomial raised to a power. However, for simpler cases like (x+3)³, repeated multiplication can be a straightforward alternative.