Write (2x+3y)^3 in expanded form
I need a little help as far as how I would begin working this problem out.
im not positively sure bc its been awhile but i think everything in the parenthesis will be raised to the 3:
2^3x^3 + 3^3y^3 giving you
8x^3 + 27y^3
To expand this you need to recall Pascal's Triangle, and use it to expand
(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
You have a=2x and b=3y, so you end up with
(2x)^3 + 3(2x)^2(3y) + 3(2x)(3y)^2 + (3y)^3
= 8x^3 + 24x^2y + 54xy^2 + 27y^3
To expand the expression (2x + 3y)^3, you can use the binomial expansion theorem. This theorem states that for any expression of the form (a + b)^n, the expanded form can be found by multiplying out the terms using the binomial coefficients and the powers of a and b.
To begin working out the problem, you can use the binomial expansion formula:
(a + b)^3 = C(3, 0) * a^3 * b^0 + C(3, 1) * a^2 * b^1 + C(3, 2) * a^1 * b^2 + C(3, 3) * a^0 * b^3
where C(n, r) represents the binomial coefficient, which is calculated using the formula C(n, r) = n! / (r! * (n - r)!), and "!" denotes factorial.
In this case, the expression is (2x + 3y)^3. So, we can substitute a = 2x and b = 3y into the formula:
(2x + 3y)^3 = C(3, 0) * (2x)^3 * (3y)^0 + C(3, 1) * (2x)^2 * (3y)^1 + C(3, 2) * (2x)^1 * (3y)^2 + C(3, 3) * (2x)^0 * (3y)^3
Simplifying each term:
= C(3, 0) * 2^3 * x^3 * 1^0 + C(3, 1) * 2^2 * x^2 * 3^1 * y^1 + C(3, 2) * 2^1 * x^1 * 3^2 * y^2 + C(3, 3) * 2^0 * x^0 * 3^3 * y^3
Next, let's calculate the binomial coefficients:
C(3, 0) = 3! / (0! * (3 - 0)!) = 1
C(3, 1) = 3! / (1! * (3 - 1)!) = 3
C(3, 2) = 3! / (2! * (3 - 2)!) = 3
C(3, 3) = 3! / (3! * (3 - 3)!) = 1
Now we can substitute these values back into the expanded form:
= 1 * 2^3 * x^3 * 1^0 + 3 * 2^2 * x^2 * 3^1 * y^1 + 3 * 2^1 * x^1 * 3^2 * y^2 + 1 * 2^0 * x^0 * 3^3 * y^3
Simplifying further:
= 8x^3 + 12x^2 * 3y + 6x * 9y^2 + 27y^3
And finally, the expanded form of (2x + 3y)^3 is:
8x^3 + 36x^2y + 54xy^2 + 27y^3.