SOLUTION: how do you solve these equations step by step? the cubed root of 625 over the cubed root of 25? the square root of 3 minus 2 times the fourth root of 3 plus 4 times the squared r

Algebra ->  Radicals -> SOLUTION: how do you solve these equations step by step? the cubed root of 625 over the cubed root of 25? the square root of 3 minus 2 times the fourth root of 3 plus 4 times the squared r      Log On


   

Question 271389: how do you solve these equations step by step?
the cubed root of 625 over the cubed root of 25?
the square root of 3 minus 2 times the fourth root of 3 plus 4 times the squared root of 3 plus 5 times the fourth root of 3?

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
First of all, these are expressions not equations. You solve equations. You simplify expressions.

root%283%2C+625%29%2Froot%283%2C+25%29
The simplest way to simplify this is to recognize that 625 can be divided evenly by 25. So we can use the property of radicals, root%28a%2C+p%29%2Froot%28a%2C+q%29+=+root%28a%2C+p%2Fq%29, to get 625 and 25 into the same fraction:
root%283%2C+625%2F25%29
Since 625/25 = 25 we get:
root%283%2C+25%29
The only other simplifying to try is to find perfect cube factors of 25. But there are none (except 1 which is pointless to factor out). So this is our simplified expression.

sqrt%283%29+-+2root%284%2C+3%29+%2B+4sqrt%283%29+%2B+5root%284%2C+3%29
The first and third terms are like terms. And the second and fourth terms are like terms so we can add and subtract them. Just like x + 4x = 5x, sqrt%283%29+%2B+4sqrt%283%29+=+5sqrt%283%29. And just like -2x + 5x = 3x, -2root%284%2C+3%29+%2B+5root%284%2C+3%29+=+3root%284%2C+3%29. So we end up with:
5sqrt%283%29+%2B+3root%284%2C+3%29