Answer to Question #125276 in Algebra for EMMANUEL

Question

Simplify 10^(1/3)n ×15^(1/2)n ×6^(1/6)n ÷45^(1/3)n

Answer 1

10^(1/3)n × 15^(1/2)n × 6^(1/6)n ÷ 45^(1/3)n

Get the factors of:

10=2×5

15=3×5

6=2×3

45=3²×5

{2^(1/3)n × 5^(1/3)n × 3^(1/2)n × 5^(1/2)n × 2^(1/6)n × 3^(1/6)n} ÷{3^(2/3)n × 5^(1/3)n}

Group common bases

{2^(1/3)n × 2^(1/6)n} × {3^(1/2)n × 3^(1/6)n ÷ 3^(2/3)n} × {5^(1/2)n × 5^(1/3)n ÷ 5^(1/3)n}

2^(1/2)n × 3⁰ × 5^(1/2)n

10^(1/2)n