%5E3.jpeg "(3t-4)^3")
Expanding (3t - 4)³
Expanding a binomial raised to a power can be done using the binomial theorem or by using repeated multiplication. Let's explore both methods to expand (3t - 4)³.
Method 1: Binomial Theorem
The binomial theorem states that for any real numbers a and b and any non-negative integer n:
(a + b)ⁿ = ∑(k=0)^n (n choose k) a^(n-k) b^k
where (n choose k) is the binomial coefficient, calculated as n!/(k!(n-k)!).
Let's apply this to our problem:
(3t - 4)³ = ∑(k=0)³ (3 choose k) (3t)^(3-k) (-4)^k
We can expand this:
- k = 0: (3 choose 0) (3t)³ (-4)⁰ = 1 * 27t³ * 1 = 27t³
- k = 1: (3 choose 1) (3t)² (-4)¹ = 3 * 9t² * -4 = -108t²
- k = 2: (3 choose 2) (3t)¹ (-4)² = 3 * 3t * 16 = 144t
- k = 3: (3 choose 3) (3t)⁰ (-4)³ = 1 * 1 * -64 = -64
Therefore, (3t - 4)³ = 27t³ - 108t² + 144t - 64
Method 2: Repeated Multiplication
We can also expand (3t - 4)³ by repeatedly multiplying:
(3t - 4)³ = (3t - 4)(3t - 4)(3t - 4)
First, multiply the first two factors:
(3t - 4)(3t - 4) = 9t² - 12t - 12t + 16 = 9t² - 24t + 16
Now, multiply this result by the third factor:
(9t² - 24t + 16)(3t - 4) = 27t³ - 36t² - 72t² + 96t + 48t - 64
Combining like terms, we get:
(3t - 4)³ = 27t³ - 108t² + 144t - 64
Conclusion
Both methods lead to the same result: (3t - 4)³ = 27t³ - 108t² + 144t - 64. Choosing the method depends on personal preference and the complexity of the problem. For simpler cases, repeated multiplication might be easier, while for larger powers, the binomial theorem offers a more structured approach.