How do you simplify #sqrt(-2)^6#?

1 Answer
Jan 15, 2017

#(sqrt(-2))^6 = -8#

Explanation:

Note that if #a# is any number and #m, n# are positive integers then:

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

#color(white)((a^m)^n) = overbrace(overbrace((axxaxx...xxa))^"m times"xxoverbrace((axxaxx...xxa))^"m times"xx...xxoverbrace((axxaxx...xxa))^"m times")^"n times"#

#color(white)((a^m)^n) = overbrace(axxaxx...xxa)^"mn times"#

#color(white)((a^m)^n) = a^(mn)#

So in our example we find:

#(sqrt(-2))^6 = (sqrt(-2))^(2*3) = (sqrt(-2)^2)^3 = (-2)^3 = -8#

#color(white)()#
Footnote

I demonstrated #(a^m)^n = a^(mn)# above since this particular "rule" fails for negative or complex values of #a# if #m, n# are fractional.

For example:

#-1 = (-1)^1 = (-1)^(3/2*2/3) != ((-1)^(3/2))^(2/3) = (-i)^(2/3) = 1/2-sqrt(3)/2i#