`a^3xxb^3` and `a^nxxb^nxxc^nxxd^n`

(a^4xxb^5xxc^7)/(c^8xxa^4xxb^3)

Simplify : a. 15^(7)-:5^(3) b. ((5^(4))^(3)xx2^(12))/(3^(12)) c. ((4^(3))^(2)xx5^(6))/(10^(6)) d. (2xx3^(4)xx2^(5))/(9xx4^(2)) e. (2^(2)xx3^(4)xx16)/(3^(2)xx32) f. (4^(5)xxa^(8)xxb^(16))/(4^(4)xxa^(7)xxb^(4))

Simplify : (3xx9^(n+1)-9xx3^(2n))/(3xx3^(2n+3)-9^(n+1))

if (9^nxx3^2xx3^n-(27)^n)/((3^3)^5xx2^3)=1/(27), find the value of n

Using laws of exponents, simplify and write the answer in exponential from: (i) a^4xxb^4 (ii) (2^(20)-:2^(15))xx2^3

Simplify and express each of the following in exponential from: (i) {(2^2)^3xx3^6}xx5^6 (ii) (a/b)^5xxb^(10)

If lim_(x to 3) (X^(n) - 3^(n))/(x - 3) = 405 and n in N Find n

Let A be a set of n(geq3) distance elements. The number of triplets (x ,y ,z) of the A elements in which at least two coordinates is equal to a. ^n P_3 b. n^3-^n P_3 c. 3n^2-2n d. 3n^2-(n-1)

Let n in N, S_n=sum_(r=0)^(3n)^(3n)C_r and T_n=sum_(r=0)^n^(3n)C_(3r) , then |S_n-3T_n| equals

Simplify : (3xx27^(n+1)+9xx3^(3n-1))/(8xx3^(3n)-5xx27^(n))