Question

1) (a+b)^3 2) (x-3y)^3

Answer 1

$\displaystyle \tt 1) \: (a+b)^3=\bold{a^3+3a^2b+3ab^2+b^3}\\\displaystyle \tt 2) \: (x-3y)^3=x^3-3x^2\cdot3y+3x\cdot(3y)^2-(3y)^3=x^3-9x^2y+3x\cdot9y^2-27y^3=\bold{x^3-9x^2y+27xy^2-27y^3}$

Answer 2

To expand the given expressions, we can use the binomial expansion formula. The formula for expanding (a + b)^n is:

(a + b)^n = Σ [ n choose k ] * a^(n-k) * b^k from k = 0 to n

Where "n choose k" is the binomial coefficient, given by n! / (k! * (n-k)!).

Let's apply this formula to the given expressions:

1. (a + b)^3:

Using the formula for (a + b)^n with n = 3: (a + b)^3 = (3 choose 0) * a^3 * b^0 + (3 choose 1) * a^2 * b^1 + (3 choose 2) * a^1 * b^2 + (3 choose 3) * a^0 * b^3

Now, calculate the binomial coefficients: (3 choose 0) = 1 (3 choose 1) = 3 (3 choose 2) = 3 (3 choose 3) = 1

The expanded expression is: (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

2. (x - 3y)^3:

Using the formula for (a + b)^n with n = 3, where a = x and b = -3y: (x - 3y)^3 = (3 choose 0) * x^3 * (-3y)^0 + (3 choose 1) * x^2 * (-3y)^1 + (3 choose 2) * x^1 * (-3y)^2 + (3 choose 3) * x^0 * (-3y)^3

Now, calculate the binomial coefficients: (3 choose 0) = 1 (3 choose 1) = 3 (3 choose 2) = 3 (3 choose 3) = 1

Remember that (-3y)^0 = 1, (-3y)^1 = -3y, (-3y)^2 = 9y^2, and (-3y)^3 = -27y^3.

The expanded expression is: (x - 3y)^3 = x^3 - 3x^2y + 9xy^2 - 27y^3

So, the expanded forms of the given expressions are:

1. (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
2. (x - 3y)^3 = x^3 - 3x^2y + 9xy^2 - 27y^3