Polynomials and Exponents

nth root of a:- If a is any real number and n is a positive integer, then the nth root of a is the real number x such xⁿa =

The nth root of a denoted by 1/aⁿ or n?a

Thus, n?a = x ? 1/aⁿ = xⁿ ? xⁿ = a

(i) ?a = 1/a² is called the square-root of a;

(ii) 3?a = 1/a³ is called the cube-root of a;

(iii) 4?a = 1/a⁴ is called the 4th-root of a and so on.

Remark: For positive value of a, the value of 1/aⁿ will always e taken as positive.

Example:

(i) ?9 = 1/9² = 3, since 3² = 9;

(ii) 3?8 = 1/8³ = 2, since 2³ = 8;

(iii) 4?81 = (81)^1/4 = 3; since 3⁴ = 81

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

Thus, (– 4)^1/2is not defined

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

For any rational number p/q, we define:

p/a^q = (1/a^q)^p = (a^p)^1/q = q?a^p

Example:

(i) 3/4² = (4³)^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

Examples on Rational and Fractional

Example1:- Simplify: 5ⁿ^{+ 3} 6 × 5ⁿ^{+ 1}/ 9 × 5ⁿ 2² × 5ⁿ

Solution. We have

5ⁿ^{+ 3} 6 × 5ⁿ^{+ 1}/ 9 × 5ⁿ 2² × 5ⁿ = 5² × 5ⁿ^{+ 1} 6 × 5ⁿ^{+ 1}/ 9 × 5 4 × 5ⁿ= 25 × 5^{(n + 1)}6 × 5^{(n + 1)} 6 × 5ⁿ4 × 5ⁿ

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

Example2:- Simplify

(i) (x^b/ x^c)^a ·(x^c/ x^a)^b · (x^a/ x^b)^c (ii) (x^a/ x^b)^{(a2 + ab + b2)}· (x^b/ x^c)^{(b2 + bc + c2)} · (x^c/ x^a)^{(c2 + ca + a2)}

Solution. We have

(i) (x^b/ x^c)^a ·(x^c/ x^a)^b · (x^a/ x^b)^c

= (x^b^{– c})^a ·(x^c^{– a})^b · (x^a^{– b})^c = x^a^{(b – c)}· x^b^{(c – a)}· x^c^{(a – b)}

= x^a^{(b – c) + b (c – a) + c (a – b)}= x⁰ = 1

(ii) (x^a/ x^b)^{(a2 +ab + b2)}· (x^b/ x^c)^(b²^{+ b}^c^{+ c}²⁾·(x^c/x^a)^(c²^{+ c}^a^{+ a2)}

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

= 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)} = x⁰ = 1

Solve the Problem

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

2. 3 × (27)ⁿ^{+ 1} + 9 × 3^{(3n – 1)}/ 8 × 3³ⁿ – 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

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