Expanding (2+h)^3
I get 8h+8h^2+2h^3
but...
the answer is 8+12h+6h^2+h^3
...What am i doing wrong?
I can't say how you got it wrong, since all I see is your answer. The stated answer is correct.
The binomial theorem for power 3 is this:
To really highlight the pattern, see that that can be written as:
(That's because anything is 1 times itself, and because anything (non-zero) raised to the zero power is 1.)
The pattern is: The powers of a start at 3 and drop by 1 with each term, ending at power = 0. The powers of b start at 0 and increase by 1 with each term, ending at power = 3.
The coefficient pattern is 1, 3, 3, 1.
Notice that the sum of the powers is 3 for each term. Notice that the formula is symmetric in order (because must equal ... and it does.)
Examples:
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As was pointed out, we can only point out mistakes if you show your work.
Two ways to proceed
First method: brute force.
$(2 + h)^3 = (2 + h)^2(2 + h) = 2(2 + h)^2 + h(2 + h)^2 =$
$2(4 + 4h + h^2) + h(4 + 4h + h^2) = 8 + 8h + 2h^2 + 4h + 4h^2 + h^3 = 8 + 12h + 6h^2 + h^3.$
Brute force is not so hard with a power of 3, but it gets pretty messy with a power of 8 or 10.
Second method: memorize the binomial expansion.
$\displaystyle (a + b)^n = \sum_{i=0}^n\dbinom{n}{i}a^{(n-i)}b^i = \sum_{i=0}^n\dfrac{n!}{i! * (n - i)!}a^{(n-i)}b^i.$
In which case you get the answer almost automatically simply by plugging in:
$(2 + h)^3 = \dfrac{3!}{0! * 3!}2^{(3-0)}h^0 + \dfrac{3!}{1! * 2!}2^{(3-1)}h^1 + \dfrac{3!}{2! * 1!}2^{(3-2)}h^2 + \dfrac{3!}{3! * 0!}2^{(3-3)}h^3 \implies$
$(2 + h)^3 = 2^3 + 3 * 2^2h + 3 * 2h^2 + h^3 = 8 + 12h + 6h^2 + h^3.$