How do you simplify #\frac { 8c d ^ { 4} } { 2c d ^ { 8} \cdot c d \cdot c ^ { 6} d }#? Prealgebra Exponents, Radicals and Scientific Notation Exponents 1 Answer Shwetank Mauria Jun 2, 2017 #(8cd^4)/(2cd^8*cd*c^6d)=4/(c^7d^6)# Explanation: #(8cd^4)/(2cd^8*cd*c^6d)# = #(8cd^4)/(2cxxcxxc^6xxd^8xxd xxd)# = #(8cd^4)/(2c^((1+1+6))xxd^((8+1+1))# = #(8cd^4)/(2c^8xxd^10)# = #8/2xxc/c^8xxd^4/d^10# = #4xx1/c^((8-1))xx1/d^((10-4))# = #4xx1/c^7xx1/d^6# = #4/(c^7d^6)# Answer link Related questions How do you simplify #c^3v^9c^-1c^0#? How do you simplify #(- 1/5)^-2 + (-2)^-2#? How do you simplify #(4^6)^2 #? How do you simplify #3x^(2/3) y^(3/4) (2x^(5/3) y^(1/2))^3 #? How do you simplify #4^3ยท4^5#? How do you simplify #(5^-2)^-3#? How do you simplify and write #(-5.3)^0# with positive exponents? How do you factor #12j^2k - 36j^6k^6 + 12j^2#? How do you simplify the expression #2^5/(2^3 times 2^8)#? When can I add exponents? See all questions in Exponents Impact of this question 1435 views around the world You can reuse this answer Creative Commons License