1. Binomial Expansion

Expanding (2+h)^3

I get 8h+8h^2+2h^3

but...

...What am i doing wrong?

2. Re: Binomial Expansion

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:

$\displaystyle (a + b)^3 = a^3 + 3 a^2 b + 3 a b^2 + b^3$

To really highlight the pattern, see that that can be written as:

$\displaystyle (a + b)^3 = (1)a^3 b^0 + (3) a^2 b^1 + (3) a^1 b^2 + (1) a^0 b^3$

(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 $\displaystyle (a + b)^3$ must equal $\displaystyle (b + a)^3$... and it does.)

Examples:

$\displaystyle (a + 0)^3 = (a)^3 + 3 (a)^2 (0) + 3 (a) (0)^2 + (0)^3$ $\displaystyle = a^3 + 0 + 0 + 0$ $\displaystyle = a^3$. (Check!)

$\displaystyle (a + (-a))^3 = (a)^3 + 3 (a)^2 (-a) + 3 (a) (-a)^2 + (-a)^3$ $\displaystyle = a^3 - 3a^3 + 3a^3 - a^3$ $\displaystyle = 0$. (Check!)

$\displaystyle (x + 1)^3 = (x)^3 + 3 (x)^2 (1) + 3 (x) (1)^2 + (1)^3 = x^3 + 3x^2 + 3x + 1$.

$\displaystyle (x - 1)^3 = (x)^3 + 3 (x)^2 (-1) + 3 (x) (-1)^2 + (-1)^3 = x^3 - 3x^2 + 3x - 1$.

$\displaystyle (x + 2)^3 = (x)^3 + 3 (x)^2 (2) + 3 (x) (2)^2 + (2)^3 = x^3 + 6 x^2 + 12x + 8$.

$\displaystyle (x - 2)^3 = (x)^3 + 3 (x)^2 (-2) + 3 (x) (-2)^2 + (-2)^3 = x^3 - 6 x^2 + 12x - 8$.

$\displaystyle (5x - 4)^3 = (5x)^3 + 3 (5x)^2 (-4) + 3 (5x) (-4)^2 + (-4)^3$ $\displaystyle = 125x^3 - 300 x^2 + 240x - 64$.

(Check: $\displaystyle 5^3 = 125, (3)(5)^2(-4) = -300, (3)(5)(-4)^2 = 240, (-4)^3$ $\displaystyle = -64$

$\displaystyle (2 + 1)^3 = (2)^3 + 3 (2)^2 (1) + 3 (2) (1)^2 + (1)^3 = 8 + 3(4) + 6 + 1$ $\displaystyle = 8 + 12 + 7$ $\displaystyle = 20 + 7$ $\displaystyle = 27$.

(Note that $\displaystyle 3^3 = 3(3)^2 = 3(9) = 27$. Check!)

$\displaystyle (3 + 1)^3 = (3)^3 + 3 (3)^2 (1) + 3 (3) (1)^2 + (1)^3 = 27 + 3(9) + 9 + 1$ $\displaystyle = 27 + 27 + 10$ $\displaystyle = 54 + 10$ $\displaystyle = 64$.

(Note that $\displaystyle 4^3 = 4(4)^2 = 4(16) = 64$. Check!)

$\displaystyle (3 + 2)^3 = (3)^3 + 3 (3)^2 (2) + 3 (3) (2)^2 + (2)^3$ $\displaystyle = 27 + 3(9)(2) + 9(4) + 8$ $\displaystyle = 27 + 9(6) + 36 + 8$ $\displaystyle = 27 + 54 + 36 + 8$ $\displaystyle = 81 + 44$ $\displaystyle = 125$.

(Note that $\displaystyle 5^3 = 5(5)^2 = 5(25) = 125$. Check!)

3. Re: Binomial Expansion

$(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$

try again ...

4. Re: Binomial Expansion

(h)^3 + 3(h)^2(2) + 3 (h)(2)^2 + (2)^3 = H^3 + 6h^2 +12h + 8

...

Thank you!

5. Re: Binomial Expansion

Originally Posted by macewan3135
Expanding (2+h)^3

I get 8h+8h^2+2h^3

but...

...What am i doing wrong?
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:

$(2 + h)^3 = \dfrac{3!}{0! * 3!}2^{(3-0)}h^0 + \dfrac{3!}{1! * 2!}2^{(3-1)}h^1 + \dfrac{3!}{2! * 1!}2^{(3-2)}h^2 + \dfrac{3!}{3! * 0!}2^{(3-3)}h^3 \implies$

$(2 + h)^3 = 2^3 + 3 * 2^2h + 3 * 2h^2 + h^3 = 8 + 12h + 6h^2 + h^3.$