Question

How do you simplify `(-6)^0`?

Answer 1

`color(blue)(=1`

Explanation:

• As per property:
  `color(blue)(a^0= 1`

So, applying the above:

`(-6)^0 =1`

Answer 2

1

Explanation:

`(-6)^0=(-6)^(n-n)=(-6)^n/(-6)^n=1`

Answer 3

`(-6)^0 = 1`

Explanation:

If `n` is a positive integer, then for any number `a` we can define:

`a^n = overbrace(a xx a xx .. xx a)^"n times"`

This has the pleasing property that if `m` and `n` are positive integers then:

`a^m * a^n = overbrace(a xx a xx .. xx a)^"m times" xx overbrace(a xx a xx .. xx a)^"n times"`

`=overbrace(a xx a xx .. xx a)^"m + n times" = a^(m+n)`

If `a != 0` then we can also define `a^(-n) = 1/underbrace(a xx a xx .. xx a)_"n times"`

With this definition we find that `a`'s in numerators and denominators combine or cancel in such a way that:

`a^m * a^n = a^(m+n)`

for any integers `m` and `n` (positive or negative).

In particular. If `a != 0` then `a^0 = a^(1-1) = a^1 * a^(-1) = a/a = 1`

In fact, in the case `n=0` we can think of `a^0` as being an empty product of `0` copies of `a`. The empty product has the value `1` - the identity under multiplication.

`color(white)()`
In particular, if `a = -6` then `a^0 = 1`