Question 678276
To simplify {{{root(3,750)+root(3,2058)-root(3,48)}}}, factor.
Each of those numbers must have a factor that is a perfect cube.
For example, {{{48=8*6=2^3*6}}} so
{{{root(3,48)=root(3,2^3*6)=root(3,2^3)*root(3,6)=2*root(3,6)}}}
If that expression can be simplified a lot,{{{750}}} and {{{2058}}} must be {{{6}}} times the cube of something.
 
You could methodically do a prime factorization,
and maybe that's what expected,
but maybe you could just find the helpful factors any way that is faster for you.
 
MY WAY:
{{{750=3*250=6*125=6*5^3}}} and
{{{2058=2*1029=2*3*343=6*343=6*7^3}}}
So
{{{root(3,750)+root(3,2058)-root(3,48)=root(3,5^3*6)+root(3,7^3*6)-root(3,2^3*6)=5*root(3,6)+7*root(3,6)-2*root(3,6)=(5+7-2)*root(3,6)=10*root(3,6)}}}