How do you write three different expressions that can be simplified to #x^6#?

1 Answer
Jul 6, 2018

Answer:

Here are some examples...

Explanation:

You can construct some simple ones using the laws of exponents:

#x^a * x^b = x^(a+b)#

which holds for #x > 0# or for any #x# if #a, b# are non-negative integers.

For example:

#x^2 * x^4 = x^(2+4) = x^6#

You can also construct interesting expressions by taking differences of squared binomials:

#1/2((x^6+1/2)^2-(x^6-1/2)^2)#

#=1/2((x^12+x^6+1/4)-(x^12-x^6+1/4)) = x^6#

Or you can make something really nasty by finding an algebraic expression that simplifies to #6#, e.g.

#x^(root(3)(135+78sqrt(3))+root(3)(135-78sqrt(3))) = x^6#

See https://socratic.org/questions/how-do-you-simplify-root-3-135-78sqrt-3-root-3-135-78sqrt-3