# Thread: expanding brackets

1. ## expanding brackets

If you have $\displaystyle (AL)^{-1} \left[ (AL)^\frac{1}{2} + BK^\frac{1}{2} \right]^2$, is there a way to sneak the $\displaystyle (AL)^{-1}$ into the brackets? Expanding the brackets wouldn't be the end of the world, but it would be nice to know of a better way. It would be especially useful if I ever come across a situation like $\displaystyle (AL)^{-1} \left[ (AL)^\frac{1}{2} + BK^\frac{1}{2} \right]^4$

2. Originally Posted by garymarkhov
If you have $\displaystyle (AL)^{-1} \left[ (AL)^\frac{1}{2} + BK^\frac{1}{2} \right]^2$, is there a way to sneak the $\displaystyle (AL)^{-1}$ into the brackets? Expanding the brackets wouldn't be the end of the world, but it would be nice to know of a better way. It would be especially useful if I ever come across a situation like $\displaystyle (AL)^{-1} \left[ (AL)^\frac{1}{2} + BK^\frac{1}{2} \right]^4$
If you take it inside the brackets, it becomes $\displaystyle (AL)^{-1/2}$.

3. Originally Posted by mr fantastic
If you take it inside the brackets, it becomes $\displaystyle (AL)^{-1/2}$.
Super! In the last hour, my basic algebra skills have got a big boost.

4. Originally Posted by mr fantastic
If you take it inside the brackets, it becomes $\displaystyle (AL)^{-1/2}$.
Sorry Mr F, but you actually need to exand the square first.

$\displaystyle (AL)^{-1}\left[(AL)^{\frac{1}{2}} + BK^{\frac{1}{2}}\right]^2 = (AL)^{-1}\left[AL + 2(AL)^{\frac{1}{2}}BK^{\frac{1}{2}} + B^2K\right]$

$\displaystyle = 1 + 2(AL)^{-\frac{1}{2}}BK^{\frac{1}{2}} + (AL)^{-1}B^2K$.

5. Originally Posted by Prove It
Sorry Mr F, but you actually need to exand the square first.

$\displaystyle (AL)^{-1}\left[(AL)^{\frac{1}{2}} + BK^{\frac{1}{2}}\right]^2 = (AL)^{-1}\left[AL + 2(AL)^{\frac{1}{2}}BK^{\frac{1}{2}} + B^2K\right]$

$\displaystyle = 1 + 2(AL)^{-\frac{1}{2}}BK^{\frac{1}{2}} + (AL)^{-1}B^2K$.
No you don't. eg. $\displaystyle a( b + c)^2 = (a^{1/2} b + a^{1/2} c)^2$.

More generally, $\displaystyle a^m (b + c)^n = (a^{m/n} b + a^{m/n} c)^n$.

6. Originally Posted by mr fantastic
No you don't. eg. $\displaystyle a( b + c)^2 = (a^{1/2} b + a^{1/2} c)^2$.

More generally, $\displaystyle a^m (b + c)^n = (a^{m/n} b + a^{m/n} c)^n$.
Ah I see... I misread your post. Sorry.

Still, if you follow the Order of Operations, the Exponentiation should come before the Multiplication anyway...