what is the monomial roots with remainders for {57d^25e^27f^22}

Question

Solving monomial roots with remainder by simplifying the square root expression

Answer 1

57d²⁵e²⁷f²² is the expression. 57=3×19.

For square root we look at each component. 57 has only prime factors so we cannot further simplify √57.

√d²⁵ can be reduced, because 2 (radical index) is bigger than 25, so 25/2=12 remainder 1, and we get:

d¹²√d

e²⁷ can be reduced in a similar way: 27/2=13 rem 1, so we get:

e¹³√e

and f²² becomes f¹¹, because 22/2=11 rem 0.

Combine all these terms: (√57)(d¹²√d)(e¹³√e)(f¹¹)=d¹²e¹³f¹¹√(57de).