How do you solve and check for extraneous solutions in #(-2x+6)^(1/5)=(-8+10x)^(1/5)#?
Actually, your equation will not have extraneous solutions because you can have negative numbers under the radicals.
Rewrite your equation to radical form to get a better understanding of what's going on
#root(5)(-2x+6) = root(5)(-8+10x)#
Since you're dealing with odd-indexed radicals, you don't have to worry about the sign of the two expressions that are under the radicals.
Remember that when you multiply a negative number by itself an odd number of times, you still get a negative number.
So, raise both sides of the equation to the fifth power
#(root(5)(-2x+6))^5 = (root(5)(-8 + 10x))^5#
#-2x + 6 = -8 + 10x#
Now simply rearrange and solve for
#12x = 14 implies x = 14/12 = color(green)(7/6)#
You can check this solution using a calculator
#root(5)(-2 * 7/6 + 6) = 1.296744116...#
#root(5)(-8 + 10 * 7/6) = 1.296744116...#