Can this algebraic expression be simplified?

[MaxManus]

show that:
a^-a(1-a)^a + a^1-a(1-a)^a-1
can be written as:

a^-a(1-a)^a-1

The Attempt at a Solution

a^-a(1-a)^a + a^1-a(1-a)^a-1

a^-a(1-a)^a + a¹a^-a(1-a)^a(1-a)^-1

a^-a(1-a)^a( 1 + a(1-a)^-1)

 

[danago]

show that:
a^-a(1-a)^a + a^1-a(1-a)^a-1
can be written as:

a^-a(1-a)^a-1

The Attempt at a Solution

a^-a(1-a)^a + a^1-a(1-a)^a-1

a^-a(1-a)^a + a¹a^-a(1-a)^a(1-a)^-1

a^-a(1-a)^a( 1 + a(1-a)^-1)

In your third line of working, instead of taking out a factor of (1-a)^a, try taking out a factor of (1-a)^a-1

 

[MaxManus]

Thanks

 

[Mark44]

show that:
a^-a(1-a)^a + a^1-a(1-a)^a-1
can be written as:

a^-a(1-a)^a-1

The Attempt at a Solution

a^-a(1-a)^a + a^1-a(1-a)^a-1

a^-a(1-a)^a + a¹a^-a(1-a)^a(1-a)^-1

a^-a(1-a)^a( 1 + a(1-a)^-1)

You really should get into the habit of including "=" between expessions that have the same value. If you don't, it will come back to bite you.

 

[MaxManus]

a^-a(1-a)^a + a^1-a(1-a)^a-1

= a^-a(1-a)^a + a¹a^-a(1-a)^a (1-a)^-1

= a^-a(1-a)^a( 1 + a(1-a)^-1)

Like this?

 

[Mark44]

Yep, that's it.