# Thread: Expanding binomial to a fraction power

1. ## Expanding binomial to a fraction power

I have the following:

$
(x^4 - 1)^\frac{1}{3}$

I know you can make that into a root:

$
\sqrt[3]{x^4 -1}$

but is there a way to expend that out, like you would expand:

$
(x^4 - 1)^2 = x^8 - 2x^4 + 1$

thanks

2. Sure. Just do it...

(x^4 - 1)^(1/3) = (x^4)^{1/3} + (1/3)

$
(x^{4}-1)^{\frac{1}{3}} = \frac{1}{0!}(x^4)^{\frac{1}{3}}(-1)^{0} + \frac{\frac{1}{3}}{1!}(x^4)^{\frac{1}{3}-1}(-1)^{1} + \frac{\frac{1}{3} \cdot \frac{-2}{3}}{2!}(x^4)^{\frac{1}{3}-2}(-1)^{2} + ...
$

It is a wonderfully useful result.

3. Originally Posted by TKHunny
Sure. Just do it...

(x^4 - 1)^(1/3) = (x^4)^{1/3} + (1/3)

$
(x^{4}-1)^{\frac{1}{3}} = \frac{1}{0!}(x^4)^{\frac{1}{3}}(-1)^{0} + \frac{\frac{1}{3}}{1!}(x^4)^{\frac{1}{3}-1}(-1)^{1} + \frac{\frac{1}{3} \cdot \frac{-2}{3}}{2!}(x^4)^{\frac{1}{3}-2}(-1)^{2} + ...
$

It is a wonderfully useful result.
Wait, how'd you go from:

$
(x^4 - 1)^\frac{1}{3} = (x^4)^\frac{1}{3} + (\frac{1}{3})$

to all that stuff at the bottom?

4. It is only a minor generalization from the finite Binomial Expansion

$(a + b)^{n} = a^{n} + ...$ See how it starts out with the first tern to the external exponent?

$(a + b)^{n} = a^{n} + n*a^{n-1}b + ...$ Then we start reducing the exponent on the first term and increasing the exponent on the second term. The rest is just controlling the combinations.