सरल बनाएं: (i) (root(5)(a^3b^(-2)c^(-4)))/(root5(a^8b^3c^(-9)))(ii) (125)^ (1/4)(8^(1/3)+27^(1/3)...

The rationalising factor of root(7)(a^(4)b^(3)c^(5)) is (a) root(7)(a^(3)b^(4)c^(2)) (b) root(7)(a^(3)b^(4)c^(2)) (c) root(7)(a^(2)b^(3)c^(3)) (d) root(7)(a^(2)b^(4)c^(3))

Simplify: [root(3)(root(6)(5^(9)))]^(8)[root(6)(root(3)(5^(9)))]^(8)

(5(8^(1//3)+27^(1//3))^(3))^(1//4) =

Factorise : (i) 125a^(3)+(1)/(8) " " (ii) 8a^(3)-27b^(3)

If (4b^(2)+(1)/(b^(2)))=2, then (8b^(3)+(1)/(b^(3)))=? (a) 0 (b) 1 (c) 2 (d) 5

4(12)/(125)3=?1(2)/(5)(b)1(3)/(5)(c)1(4)/(5)(d)2(2)/(5)

Find the logarithms of the following numbers to the base 2: (a) root(3)(3)( b) root(2)(2)( c) (1)/(root(5)(2)) (d) (1)/(root(7)(8))

int(root(3)(x))(root(5)(1+root(3)(x^(4))))dx(i)(1+x^((3)/(4)))^((6)/(5))+C(ii)(1+x^((6)/(3)))^((6)/(5))+C(iii)(5)/(18)(1+x^((4)/(3)))^((6)/(5))+C (iv) (1)/(6)(1+x^((4)/(3)))^(6)+C

Factorise (i) 8p^3+(12)/(5)p^2+(6)/(25)p+(1)/(125) (ii) 27p^3-(9)/(2)p^2+(1)/(4)p-(1)/(216)

(root(4)9+root(4)4)^(1//3)(root(4)9-root(4)4)^(1//3)=