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**nth root of a:- **If

The *n*th root of *a* denoted by 1/*a ^{n}* or

Thus, *n*?*a* = *x* ? 1/*a ^{n}* =

(i) ?*a* = 1/*a*^{2} is called the **square-root **of *a*;

(ii) 3?*a* = 1/*a*^{3} is called the **cube-root **of *a*;

(iii) 4?*a* = 1/*a*^{4} is called the **4th-root **of *a* and so on.

Remark: For positive value of *a*, the value of 1/*a ^{n}* will always e taken as positive.

Example:** **

(i) ?9 = 1/9^{2} = 3, since 3^{2} = 9;

(ii) 3?8 = 1/8^{3} = 2, since 2^{3} = 8;

(iii) 4?81 = (81)^{1/4 } = 3; since 3^{4} = 81

Remark: If *a* is a negative real number and *n* is an even positive integer, then 1/*a ^{n}* is not defined.

Thus, (– 4)^{1/2 }is not defined

Similarly, (– 9)^{1/4}, (– 64)^{1/6} are not defined

For any rational number *p/q*, we define:

*p*/*a ^{q}* = (1/

Example:

(i) 3/4^{2} = (4^{3})^{1/2 }= (64)^{1/2} = 8

(ii) (– 8) ^{2/3} = [ (– 8)^{2 }]^{1/3} = (64)^{1/3} = 4

(iii) (– 4)^{3/2} = [ (– 4)^{3} ]^{1/2} = (– 64)^{1/2}, which is not defined

**Example1:- **Simplify: 5^{n}^{ + 3} 6 × 5^{n}^{ + 1}/ 9 × 5* ^{n}* 2

**Solution. **We have

5^{n}^{ + 3} 6 × 5^{n}^{ + 1}/ 9 × 5* ^{n}* 2

= 5^{(n + 1)} × (25 6)/ 5^{n}^{ }× (9 4) = 19 × 5^{n}^{ + 1}/ 5 × 5* ^{n}* = 19 × 5

**Example2:- **Simplify

(i) (*x ^{b}*/

** **

**Solution. **We have

(i) (*x ^{b}*/

= (*x ^{b}*

= *x ^{a}*

^{ }(ii) (*x ^{a}*/

^{ }= (*x ^{a}*

= *x*^{(a – b) (a2 + ab + b2)} · *x*^{(b – c) (b2 + bc + c2)} · *x*^{(c – a) (c2 + ca + a2)}

= *x*^{(a3 – b3)} · *x*^{(b3 – c3)} · *x*^{(c3 – a3)} · = *x*^{(a3 – b3) + (b3 – c3) + (c3 – a3) } = *x*^{0}* *= 1

1. 5^{2(n + 6)} × (25)^{ – 7 + 2n}/(125)^{2n}

2. 3 × (27)^{n}^{ + 1} + 9 × 3^{(3n – 1)}/ 8 × 3^{3n} – 5 × (27)^{n}

3. If *abc* = 1, prove that: 1/1 + *a* + *b *^{– 1} + 1/1 + *b* + *c *^{– 1} + 1/1 + *c* + *a *^{– 1} = 1

4. If *a, b, c* are positive real numbers, show that:

?*a*^{ – 1}*b*^{ }· ?*b*^{ – 1}*c*^{ }· ?*c*^{ – 1}*a*^{ }= 1 ^{ } ^{ } ^{ }