Use the Binomial Theorem to expand the binomial:
(d – 4b)^3
d^3 + 3d^2(-4b) + 3d(-4b)^2 + (-4b)^3
= d^3 - 12d^2b + 48db^2 - 64b^3
is this right?
^ yes
The Binomial Theorem Quick Check:
1. A. or the one that has "-1,458d + 729" at the end of it.
2. B. or the one that has "+243v^5" at the end of it.
3. D. or "-64b^3"
100% 3/3
Ah, the Binomial Theorem, how much binomial fun is in store for us! Let's expand the binomial (d – 4b) raised to the power of 3:
We start with the formula of the Binomial Theorem:
(a + b)^n = C(n,0)a^n b^0 + C(n,1)a^(n-1) b^1 + ... + C(n,n-1)a^1 b^(n-1) + C(n,n)a^0 b^n
Here, a = d, b = -4b, and n = 3. And don't panic, we need to deal with those negative signs cautiously!
Now, let's plug in the values and have some binomial fun:
(d – 4b)^3 = C(3,0)d^3(-4b)^0 + C(3,1)d^2(-4b)^1 + C(3,2)d^1(-4b)^2 + C(3,3)d^0(-4b)^3
Don't worry, I'll simplify it for you, my mathematically-inclined friend:
(d – 4b)^3 = d^3 + 3d^2(-4b) + 3d(-4b)^2 + (-4b)^3
Now, to embrace our inner mathematician, let's simplify even further:
(d – 4b)^3 = d^3 - 12d^2b + 48db^2 - 64b^3
And there you have it, the expanded form of (d – 4b) raised to the power of 3. Math can be quite a clown sometimes, turning numbers into silly expressions!
To expand (d - 4b)^3 using the Binomial Theorem, you can multiply out each term using the formula:
(a + b)^n = C(n, 0)a^n b^0 + C(n, 1)a^(n-1) b^1 + C(n, 2)a^(n-2) b^2 + ... + C(n, n-1)a^1 b^(n-1) + C(n, n)a^0 b^n
In our case, the binomial is (d - 4b), and n is 3. Therefore, we have:
(d - 4b)^3 = C(3, 0)d^3 (-4b)^0 + C(3, 1)d^2 (-4b)^1 + C(3, 2)d^1 (-4b)^2 + C(3, 3)d^0 (-4b)^3
Now, let's calculate each term step by step:
C(3, 0) = 1
C(3, 1) = 3
C(3, 2) = 3
C(3, 3) = 1
(d - 4b)^3 = 1(d^3)(-4b)^0 + 3(d^2)(-4b)^1 + 3(d^1)(-4b)^2 + 1(d^0)(-4b)^3
Simplifying further:
(d - 4b)^3 = d^3 + 3d^2(-4b) + 3d(-4b)^2 + (-4b)^3
(d - 4b)^3 = d^3 - 12d^2 b + 48db^2 - 64b^3
So, the binomial (d - 4b)^3 can be expanded to d^3 - 12d^2 b + 48db^2 - 64b^3 using the Binomial Theorem.