By way of introduction, you might want to read my short article here<\/a> on perfect cubes and cube roots.<\/p>\n Of course, 3<\/sup>√(8)=2 because (2)(2)(2)=2^3=8. Note that 2^3 is said as “two cubed”.<\/p>\n What about 3<\/sup>√(40)? There is no whole number x such that x^3=40. Yet, 3<\/sup>√(40) can be simplified.<\/p>\n Step 1: Find the factor of 40 that you can take the cube root of (8 in this case).<\/p>\n Step 2: Rewrite 3<\/sup>√(40) as 3<\/sup>√(8)√3<\/sup>(5).<\/p>\n Step 3: Replace 3<\/sup>√(8)with 2.<\/p>\n We arrive at<\/p>\n 3<\/sup>√(40)=3<\/sup>√(8)3<\/sup>√(5)=2(3<\/sup>√(5))<\/p>\n The trick becomes identifying which perfect cube can be factored from the number.<\/p>\n Example:<\/strong> Simplify 3<\/sup>√(192)<\/p>\n Solution:<\/p>\n In this case, realize that 192=(64)(3). Therefore,<\/p>\n 3<\/sup>√(192)=3<\/sup>√(64)3<\/sup>√(3)<\/p>\n 3<\/sup>√(64)=4. Therefore,<\/p>\n 3<\/sup>√(192)=3<\/sup>√(64)3<\/sup>√(3)=4(3<\/sup>√(3))<\/p>\n How we discover 64 as the number to use, and more, will be explored in a post coming soon:)<\/p>\n