How do you simplify the following problem?

#root(3)24+root(3)81#

1 Answer
Mar 9, 2018

# "The answer is:" \qquad \qquad \qquad root{3}{24} + root{3}{81} \qquad = \qquad5 root{3}{3}. #

Explanation:

# "We can do this using a property or two of radicals ... " #

# "We are given ... " #

# \qquad root{3}{24} + root{3}{81} \qquad = \qquad root{3}{8 cdot 3} + root{3}{27 cdot 3} #

# \qquad \ color{blue}{ "now insert mental emphasis" \qquad rarr } #

# = \ [ root{3}{ \overbrace{ color{red}{8} }^{ " " 8 \ "is a perfect cube" } cdot 3 } \ ] \qquad + \qquad [ root{3}{\overbrace{ color{red}{27} }^ { " " 27 \ "is a perfect cube" } cdot 3 } \ \ ] #

# \qquad \ color{blue}{ "now use a basic property of radicals:" \qquad root{n}{a cdot b} \ = \ root{n}{a} cdot root{n}{b} \quad rarr } #

# = \ [ root{3}{ \overbrace{ color{red}{8} }^{ " " 8 \ "is a perfect cube" } } cdot root{3}{3} ] \quad + \quad [ root{3}{\overbrace{ color{red}{27} }^ { " " 27 \ "is a perfect cube" } } cdot root{3}{3} ] #

# \qquad \ color{blue}{ "now remove some of the inner emphasis" \qquad rarr } #

# = \ [ root{3}{ color{red}{8} } cdot root{3}{3} \ ] \quad + \quad [ root{3}{ color{red}{27} } cdot root{3}{3} \ ] #

# \qquad \ color{blue}{ "now use basic numerical facts (about 8, and 27)" \qquad rarr } #

# = \ [ color{red}{2} cdot root{3}{3} \ ] \quad + \quad [ color{red}{3} cdot root{3}{3} \ ] #

# \qquad \ color{blue}{ "now reset to a new emphasis" \qquad rarr } #

# = \ [ 2 cdot color{red}{ root{3}{3} } \ ] \quad + \quad [ 3 cdot color{red}{ root{3}{3} } \ ] #

# \qquad \ color{blue}{ "now factor out a common factor:" \qquad color{red}{ root{3}{3} } \qquad rarr } #

# = \ [ 2 + 3 ] cdot color{red}{ root{3}{3} } #

# \qquad \ color{blue}{ "now basic arithmetic, then remove emphasis & finish !!! " \quad rarr } #

# = \ [ 5 ] cdot color{red}{ root{3}{3} } #

# = \ 5 root{3}{3}. #

# "This is our answer !!" #

#"So, summarizing, we have:"#

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad root{3}{24} + root{3}{81} \qquad = \qquad5 root{3}{3}. #