# expanding brackets

• Jan 8th 2010, 05:54 PM
garymarkhov
expanding brackets
If you have $(AL)^{-1} \left[ (AL)^\frac{1}{2} + BK^\frac{1}{2} \right]^2$, is there a way to sneak the $(AL)^{-1}$ 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 $(AL)^{-1} \left[ (AL)^\frac{1}{2} + BK^\frac{1}{2} \right]^4$
• Jan 8th 2010, 05:58 PM
mr fantastic
Quote:

Originally Posted by garymarkhov
If you have $(AL)^{-1} \left[ (AL)^\frac{1}{2} + BK^\frac{1}{2} \right]^2$, is there a way to sneak the $(AL)^{-1}$ 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 $(AL)^{-1} \left[ (AL)^\frac{1}{2} + BK^\frac{1}{2} \right]^4$

If you take it inside the brackets, it becomes $(AL)^{-1/2}$.
• Jan 8th 2010, 06:12 PM
garymarkhov
Quote:

Originally Posted by mr fantastic
If you take it inside the brackets, it becomes $(AL)^{-1/2}$.

Super! In the last hour, my basic algebra skills have got a big boost.
• Jan 8th 2010, 06:25 PM
Prove It
Quote:

Originally Posted by mr fantastic
If you take it inside the brackets, it becomes $(AL)^{-1/2}$.

Sorry Mr F, but you actually need to exand the square first.

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

$= 1 + 2(AL)^{-\frac{1}{2}}BK^{\frac{1}{2}} + (AL)^{-1}B^2K$.
• Jan 8th 2010, 06:28 PM
mr fantastic
Quote:

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

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

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

No you don't. eg. $a( b + c)^2 = (a^{1/2} b + a^{1/2} c)^2$.

More generally, $a^m (b + c)^n = (a^{m/n} b + a^{m/n} c)^n$.
• Jan 8th 2010, 06:30 PM
Prove It
Quote:

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

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