%5E3-expansion.jpeg "(a+b+c+d)^3 Expansion")
Expanding (a + b + c + d)^3
The expansion of (a + b + c + d)^3 can be a daunting task, but it can be tackled systematically using the multinomial theorem.
Understanding the Multinomial Theorem
The multinomial theorem states that for any positive integer n and any non-negative integers k<sub>1</sub>, k<sub>2</sub>, ..., k<sub>r</sub> such that k<sub>1</sub> + k<sub>2</sub> + ... + k<sub>r</sub> = n, the following expansion holds:
(x<sub>1</sub> + x<sub>2</sub> + ... + x<sub>r</sub>)<sup>n</sup> = Σ (n! / (k<sub>1</sub>! k<sub>2</sub>! ... k<sub>r</sub>!)) * x<sub>1</sub><sup>k<sub>1</sub></sup> * x<sub>2</sub><sup>k<sub>2</sub></sup> * ... * x<sub>r</sub><sup>k<sub>r</sub></sup>
where the summation is taken over all possible combinations of non-negative integers k<sub>1</sub>, k<sub>2</sub>, ..., k<sub>r</sub> that satisfy k<sub>1</sub> + k<sub>2</sub> + ... + k<sub>r</sub> = n.
Applying the Multinomial Theorem to (a + b + c + d)^3
In our case, we have n = 3, and r = 4 (representing the four variables a, b, c, and d). We need to find all possible combinations of k<sub>1</sub>, k<sub>2</sub>, k<sub>3</sub>, and k<sub>4</sub> such that k<sub>1</sub> + k<sub>2</sub> + k<sub>3</sub> + k<sub>4</sub> = 3.
Here are the combinations:
- (3, 0, 0, 0): This corresponds to the term a<sup>3</sup>
- (2, 1, 0, 0): This corresponds to the term a<sup>2</sup>b
- (2, 0, 1, 0): This corresponds to the term a<sup>2</sup>c
- (2, 0, 0, 1): This corresponds to the term a<sup>2</sup>d
- (1, 2, 0, 0): This corresponds to the term ab<sup>2</sup>
- (1, 1, 1, 0): This corresponds to the term abc
- (1, 1, 0, 1): This corresponds to the term abd
- (1, 0, 2, 0): This corresponds to the term ac<sup>2</sup>
- (1, 0, 1, 1): This corresponds to the term acd
- (1, 0, 0, 2): This corresponds to the term ad<sup>2</sup>
- (0, 3, 0, 0): This corresponds to the term b<sup>3</sup>
- (0, 2, 1, 0): This corresponds to the term b<sup>2</sup>c
- (0, 2, 0, 1): This corresponds to the term b<sup>2</sup>d
- (0, 1, 2, 0): This corresponds to the term bc<sup>2</sup>
- (0, 1, 1, 1): This corresponds to the term bcd
- (0, 1, 0, 2): This corresponds to the term bd<sup>2</sup>
- (0, 0, 3, 0): This corresponds to the term c<sup>3</sup>
- (0, 0, 2, 1): This corresponds to the term c<sup>2</sup>d
- (0, 0, 1, 2): This corresponds to the term cd<sup>2</sup>
- (0, 0, 0, 3): This corresponds to the term d<sup>3</sup>
Calculating the Coefficients
Now we need to calculate the coefficients for each term using the multinomial theorem:
- (3, 0, 0, 0): (3! / (3! 0! 0! 0!)) * a<sup>3</sup> = a<sup>3</sup>
- (2, 1, 0, 0): (3! / (2! 1! 0! 0!)) * a<sup>2</sup>b = 3a<sup>2</sup>b
- (2, 0, 1, 0): (3! / (2! 0! 1! 0!)) * a<sup>2</sup>c = 3a<sup>2</sup>c
- (2, 0, 0, 1): (3! / (2! 0! 0! 1!)) * a<sup>2</sup>d = 3a<sup>2</sup>d
- (1, 2, 0, 0): (3! / (1! 2! 0! 0!)) * ab<sup>2</sup> = 3ab<sup>2</sup>
- (1, 1, 1, 0): (3! / (1! 1! 1! 0!)) * abc = 6abc
- (1, 1, 0, 1): (3! / (1! 1! 0! 1!)) * abd = 6abd
- (1, 0, 2, 0): (3! / (1! 0! 2! 0!)) * ac<sup>2</sup> = 3ac<sup>2</sup>
- (1, 0, 1, 1): (3! / (1! 0! 1! 1!)) * acd = 6acd
- (1, 0, 0, 2): (3! / (1! 0! 0! 2!)) * ad<sup>2</sup> = 3ad<sup>2</sup>
- (0, 3, 0, 0): (3! / (0! 3! 0! 0!)) * b<sup>3</sup> = b<sup>3</sup>
- (0, 2, 1, 0): (3! / (0! 2! 1! 0!)) * b<sup>2</sup>c = 3b<sup>2</sup>c
- (0, 2, 0, 1): (3! / (0! 2! 0! 1!)) * b<sup>2</sup>d = 3b<sup>2</sup>d
- (0, 1, 2, 0): (3! / (0! 1! 2! 0!)) * bc<sup>2</sup> = 3bc<sup>2</sup>
- (0, 1, 1, 1): (3! / (0! 1! 1! 1!)) * bcd = 6bcd
- (0, 1, 0, 2): (3! / (0! 1! 0! 2!)) * bd<sup>2</sup> = 3bd<sup>2</sup>
- (0, 0, 3, 0): (3! / (0! 0! 3! 0!)) * c<sup>3</sup> = c<sup>3</sup>
- (0, 0, 2, 1): (3! / (0! 0! 2! 1!)) * c<sup>2</sup>d = 3c<sup>2</sup>d
- (0, 0, 1, 2): (3! / (0! 0! 1! 2!)) * cd<sup>2</sup> = 3cd<sup>2</sup>
- (0, 0, 0, 3): (3! / (0! 0! 0! 3!)) * d<sup>3</sup> = d<sup>3</sup>
The Final Expansion
Therefore, the complete expansion of (a + b + c + d)<sup>3</sup> is:
(a + b + c + d)<sup>3</sup> = a<sup>3</sup> + 3a<sup>2</sup>b + 3a<sup>2</sup>c + 3a<sup>2</sup>d + 3ab<sup>2</sup> + 6abc + 6abd + 3ac<sup>2</sup> + 6acd + 3ad<sup>2</sup> + b<sup>3</sup> + 3b<sup>2</sup>c + 3b<sup>2</sup>d + 3bc<sup>2</sup> + 6bcd + 3bd<sup>2</sup> + c<sup>3</sup> + 3c<sup>2</sup>d + 3cd<sup>2</sup> + d<sup>3</sup>
Conclusion
While the process of expanding (a + b + c + d)<sup>3</sup> may seem complex at first, understanding and applying the multinomial theorem allows you to systematically calculate each term and coefficient, leading to the complete expansion. This powerful tool can be used to expand expressions with any number of variables raised to any power.