SlideShare a Scribd company logo
1 of 5
Download to read offline
Free GK Alerts- JOIN OnlineGK to 9870807070


                     9. SURDS AND INDICES
                         IMPORTANT FACTS AND FORMULA




                                                                    K
1. LAWS OF INDICES:

    (i) am x an = am + n




                                                                  eG
    (ii) am / an = am-n
    (iii)   (am)n = amn
    (iv)    (ab)n = anbn
    (v) ( a/ b )n = ( an / bn )
    (vi)    a0 = 1


2. SURDS: Let a be a rational number and n be a positive integer such that a1/n = nsqrt(a)


3. LAWS OF SURDS:

(i) n√a = a1/2
                                               in
  is irrational. Then nsqrt(a) is called a surd of order n.
                             nl
(ii) n √ab = n √a * n √b
(iii) n √a/b = n √a / n √b
(iv) (n √a)n = a
(v) m√(n√(a)) = mn√(a)
(vi) (n√a)m = n√am
   eO

I SOLVED EXAMPLES


Ex. 1. Simplify : (i) (27)2/3     (ii) (1024)-4/5 (iii)( 8 / 125 )-4/3

Sol .   (i) (27)2/3 = (33)2/3 = 3( 3 * ( 2/ 3)) = 32 = 9
Th



        (ii) (1024)-4/5 = (45)-4/5 = 4 { 5 * ( (-4) / 5 )} = 4-4 = 1 / 44 = 1 / 256
        (iii) ( 8 / 125 )-4/3 = {(2/5)3}-4/3 = (2/5){ 3 * ( -4/3)} = ( 2 / 5 )-4 = ( 5 / 2 )4 = 54 / 24 = 625 / 16


Ex. 2. Evaluate: (i) (.00032)3/5            (ii)l (256)0.16 x (16)0.18.

Sol. (i) (0.00032)3/5 = ( 32 / 100000 )3/5. = (25 / 105)3/5 = {( 2 / 10 )5}3/5 = ( 1 / 5 )(5 * 3 / 5) =
(1/5)3 = 1 / 125
       (ii) (256)0. 16 * (16)0. 18 = {(16)2}0. 16 * (16)0. 18 = (16)(2 * 0. 16) * (16)0. 18
=(16)0.32 * (16)0.18 = (16)(0.32+0.18) = (16)0.5 = (16)1/2 = 4.
 196



Ex. 3. What is the quotient when (x-1 - 1) is divided by (x - 1) ?




                                                                     K
Sol.  x-1 -1 = (1/x)-1 = _1 -x * 1 = -1
      x-1       x-1        x (x - 1)  x
Hence, the required quotient is -1/x

Ex. 4. If 2x - 1 + 2x + 1 = 1280, then find the value of x.




                                                                   eG
Sol. 2x - 1 + 2X+ 1 = 1280  2x-1 (1 +22) = 1280

                                 2x-1 = 1280 / 5 = 256 = 28  x -1 = 8  x = 9.


Hence, x = 9.



                                                in
Ex. 5. Find the value of [ 5 ( 81/3 + 271/3)3]1/ 4
                             nl
Sol.    [ 5 ( 81/3 + 271/3)3]1/ 4 = [ 5 { (23)1/3 + (33)1/3}3]1/ 4 = [ 5 { (23 * 1/3)1/3 + (33 *1/3 )1/3}3]1/ 4

                                   = {5(2+3)3}1/4 = (5 * 53)1/ 4 =5(4 * 1/ 4) = 51 = 5.
   eO

Ex. 6. Find the Value of {(16)3/2 + (16)-3/2}


Sol.   [(16)3/2 +(16)-3/2 = (42)3/2 +(42)-3/2 = 4(2 * 3/2) + 4{ 2* (-3/2)}
                        = 43 + 4-3 = 43 + (1/43) = ( 64 + ( 1/64)) = 4097/64.

Ex. 7. If (1/5)3y = 0.008, then find the value of(0.25)y.
Th



Sol. (1/5)3y = 0.008 = 8/1000 = 1/125 = (1/5)3  3y = 3  Y = 1.
        (0.25)y = (0.25)1 = 0.25.
Ex. 8. Find the value of               (243)n/5  32n + 1
                                        9n  3n -1 .

Sol. (243)n/5 x32n+l = 3 (5 * n/5)  32n+l _ = 3n 32n+1
     (32)n  3n - 1   32n  3n - 1             32n  3n-l




                                                                              K
                                 = 3n + (2n + 1) = 3(3n+1) = 3(3n+l)-(3n-l) = 32 = 9.
                                    32n+n-1        3(3n-1)

Ex. 9. Find the value Of (21/4-1)(23/4+21/2+21/4+1)




                                                                            eG
Sol.
          Putting 21/4 = x, we get :

          (21/4-1) (23/4+21/2+21/4+1)=(x-1)(x3+x2+x+1) , where x = 21/4
                                     =(x-1)[x2(x+1)+(x+1)]
                                     =(x-1)(x+1)(x2+1) = (x2-1)(x2+1)


Ex. 10. Find the value of 62/3  3√67
                              3
                                √66
                                                           in
                                     =(x4-1) = [(21/4)4-1] = [2(1/4*4) –1] = (2-1) = 1.
                                    nl
Sol.       62/3  3√67 = 62/3  (67)1/3 = 62/3  6(7 * 1/3) = 62/3  6(7/3)
               3
                 √66        (66)1/3          6(6 * 1/3)           62

                       =62/3  6((7/3)-2) = 62/3  61/3 = 61 = 6.
Ex. 11. If x= y , y=z and z=xc,then find the value of abc.
                    a        b
    eO

Sol.     z1= xc =(ya)c [since x= ya]
           =y(ac) = (zb)ac [since y=zb]
           =zb(ac)= zabc
          abc = 1.

Ex. 12. Simplify [(xa / xb)^(a2+b2+ab)] * [(xb / xc )^ b2+c2+bc)] * [(xc/xa)^(c2+a2+ca)]
Th



Sol.
Given Expression
  = [{x(o - b)}^(a2 + b2 + ob)].['(x(b - c)}^ (b2 + c2 + bc)].['(x(c - a)}^(c2 + a2 + ca])
= [x(a - b)(a2 + b2 + ab) . x(b - c) (b2 +c2+ bc).x(c- a) (c2 + a2 + ca)]
= [x^(a3-b3)].[x^(b3-e3)].[x^(c3-a3)] = x^(a3-b3+b3-c3+c3-a3) = x0 = 1.
Ex. 13. Which is larger √2 or 3√3 ?




                                                                K
Sol. Given surds are of order 2 and 3. Their L.C.M. is 6. Changing each to a surd of order 6, we
get:
                  √2 = 21/2 = 2((1/2)*(3/2)) =23/6 = 81/6 = 6√8
                  3
                   √3= 31/3 = 3((1/3)*(2/2)) = 32/6 = (32)1/6 = (9)1/6 = 6√9.




                                                              eG
Clearly, √9 > √8 and hence 3√3 > √2.
         6     6



Ex. 14. Find the largest from among 4√6, √2 and 3√4.
Sol. Given surds are of order 4, 2 and 3 respectively. Their L.C,M, is 12, Changing each to a surd
of order 12, we get:

4
 √6 = 61/4 = 6((1/4)*(3/3)) = 63/12 = (63)1/12 = (216)1/12.

3
                                           in
√2 = 21/2 = 2((1/2)*(6/6)) = 26/12 = (26)1/12 = (64)1/12.
 √4 = 41/3 = 4((1/3)*(4/4)) = 44/12 = (44)1/12 = (256)1/12.

Clearly, (256)1/12 > (216)1/12 > (64)1/12
                          nl
Largest one is (256)1/12. i.e. 3√4 .
    eO
Th
.
Th
  eO
     nl
       in
          eG
            K

More Related Content

What's hot

Matematika Kalkulus ( Limit )
Matematika Kalkulus ( Limit )Matematika Kalkulus ( Limit )
Matematika Kalkulus ( Limit )fdjouhana
 
Who wants to pass this course
Who wants to pass this courseWho wants to pass this course
Who wants to pass this courseadvancedfunctions
 
Special products and factorization / algebra
Special products and factorization / algebraSpecial products and factorization / algebra
Special products and factorization / algebraindianeducation
 
Complex Numbers 1 - Math Academy - JC H2 maths A levels
Complex Numbers 1 - Math Academy - JC H2 maths A levelsComplex Numbers 1 - Math Academy - JC H2 maths A levels
Complex Numbers 1 - Math Academy - JC H2 maths A levelsMath Academy Singapore
 
Module 10 Topic 3 factoring perfect square & difference of square
Module 10 Topic 3   factoring perfect square & difference of squareModule 10 Topic 3   factoring perfect square & difference of square
Module 10 Topic 3 factoring perfect square & difference of squareLori Rapp
 
Ισοδύναμα κλάσματα
Ισοδύναμα κλάσματαΙσοδύναμα κλάσματα
Ισοδύναμα κλάσματαteaghet
 
Integration with limits
Integration with limitsIntegration with limits
Integration with limitsShaun Wilson
 
Digital textbook -EXPONENTS AND POWERS
Digital textbook -EXPONENTS AND POWERSDigital textbook -EXPONENTS AND POWERS
Digital textbook -EXPONENTS AND POWERSGANESHKRISHNANG
 
Distance book title
Distance  book titleDistance  book title
Distance book titleNadeem Uddin
 

What's hot (18)

Chapter 2(limits)
Chapter 2(limits)Chapter 2(limits)
Chapter 2(limits)
 
Matematika Kalkulus ( Limit )
Matematika Kalkulus ( Limit )Matematika Kalkulus ( Limit )
Matematika Kalkulus ( Limit )
 
F2 algebraic expression ii
F2   algebraic expression iiF2   algebraic expression ii
F2 algebraic expression ii
 
Who wants to pass this course
Who wants to pass this courseWho wants to pass this course
Who wants to pass this course
 
Leidy rivadeneira deber_1
Leidy rivadeneira deber_1Leidy rivadeneira deber_1
Leidy rivadeneira deber_1
 
Indeks(t3)
Indeks(t3)Indeks(t3)
Indeks(t3)
 
Alg2 lesson 10-3
Alg2 lesson 10-3Alg2 lesson 10-3
Alg2 lesson 10-3
 
Gr 11 equations
Gr 11   equationsGr 11   equations
Gr 11 equations
 
Special products and factorization / algebra
Special products and factorization / algebraSpecial products and factorization / algebra
Special products and factorization / algebra
 
Complex Numbers 1 - Math Academy - JC H2 maths A levels
Complex Numbers 1 - Math Academy - JC H2 maths A levelsComplex Numbers 1 - Math Academy - JC H2 maths A levels
Complex Numbers 1 - Math Academy - JC H2 maths A levels
 
Module 10 Topic 3 factoring perfect square & difference of square
Module 10 Topic 3   factoring perfect square & difference of squareModule 10 Topic 3   factoring perfect square & difference of square
Module 10 Topic 3 factoring perfect square & difference of square
 
Ισοδύναμα κλάσματα
Ισοδύναμα κλάσματαΙσοδύναμα κλάσματα
Ισοδύναμα κλάσματα
 
Integration with limits
Integration with limitsIntegration with limits
Integration with limits
 
Distance formula
Distance formulaDistance formula
Distance formula
 
Digital textbook -EXPONENTS AND POWERS
Digital textbook -EXPONENTS AND POWERSDigital textbook -EXPONENTS AND POWERS
Digital textbook -EXPONENTS AND POWERS
 
Distance book title
Distance  book titleDistance  book title
Distance book title
 
Special product
Special productSpecial product
Special product
 
Difference of squares
Difference of squaresDifference of squares
Difference of squares
 

Similar to 9 chap

RS Agarwal Quantitative Aptitude - 9 chap
RS Agarwal Quantitative Aptitude - 9 chapRS Agarwal Quantitative Aptitude - 9 chap
RS Agarwal Quantitative Aptitude - 9 chapVinoth Kumar.K
 
RS Agarwal Quantitative Aptitude - 4 chap
RS Agarwal Quantitative Aptitude - 4 chapRS Agarwal Quantitative Aptitude - 4 chap
RS Agarwal Quantitative Aptitude - 4 chapVinoth Kumar.K
 
Stacks image 1721_36
Stacks image 1721_36Stacks image 1721_36
Stacks image 1721_36Dreams4school
 
RS Agarwal Quantitative Aptitude - 5 chap
RS Agarwal Quantitative Aptitude - 5 chapRS Agarwal Quantitative Aptitude - 5 chap
RS Agarwal Quantitative Aptitude - 5 chapVinoth Kumar.K
 
5.1 sequences and summation notation t
5.1 sequences and summation notation t5.1 sequences and summation notation t
5.1 sequences and summation notation tmath260
 
19 trig substitutions-x
19 trig substitutions-x19 trig substitutions-x
19 trig substitutions-xmath266
 
1586746631GAMMA BETA FUNCTIONS.pdf
1586746631GAMMA BETA FUNCTIONS.pdf1586746631GAMMA BETA FUNCTIONS.pdf
1586746631GAMMA BETA FUNCTIONS.pdfFighting2
 
Algebra Revision.ppt
Algebra Revision.pptAlgebra Revision.ppt
Algebra Revision.pptAaronChi5
 
Cbse Class 12 Maths Sample Paper 2013 Model 3
Cbse Class 12 Maths Sample Paper 2013 Model 3Cbse Class 12 Maths Sample Paper 2013 Model 3
Cbse Class 12 Maths Sample Paper 2013 Model 3Sunaina Rawat
 
The solution of problem
The solution of problemThe solution of problem
The solution of problemnoviannurf
 
1.3 solving equations t
1.3 solving equations t1.3 solving equations t
1.3 solving equations tmath260
 
resposta do capitulo 15
resposta do capitulo 15resposta do capitulo 15
resposta do capitulo 15silvio_sas
 

Similar to 9 chap (20)

RS Agarwal Quantitative Aptitude - 9 chap
RS Agarwal Quantitative Aptitude - 9 chapRS Agarwal Quantitative Aptitude - 9 chap
RS Agarwal Quantitative Aptitude - 9 chap
 
9. surds & indices
9. surds & indices9. surds & indices
9. surds & indices
 
RS Agarwal Quantitative Aptitude - 4 chap
RS Agarwal Quantitative Aptitude - 4 chapRS Agarwal Quantitative Aptitude - 4 chap
RS Agarwal Quantitative Aptitude - 4 chap
 
4 chap
4 chap4 chap
4 chap
 
Stacks image 1721_36
Stacks image 1721_36Stacks image 1721_36
Stacks image 1721_36
 
New stack
New stackNew stack
New stack
 
RS Agarwal Quantitative Aptitude - 5 chap
RS Agarwal Quantitative Aptitude - 5 chapRS Agarwal Quantitative Aptitude - 5 chap
RS Agarwal Quantitative Aptitude - 5 chap
 
4. simplifications
4. simplifications4. simplifications
4. simplifications
 
5.1 sequences and summation notation t
5.1 sequences and summation notation t5.1 sequences and summation notation t
5.1 sequences and summation notation t
 
19 trig substitutions-x
19 trig substitutions-x19 trig substitutions-x
19 trig substitutions-x
 
1586746631GAMMA BETA FUNCTIONS.pdf
1586746631GAMMA BETA FUNCTIONS.pdf1586746631GAMMA BETA FUNCTIONS.pdf
1586746631GAMMA BETA FUNCTIONS.pdf
 
Algebra Revision.ppt
Algebra Revision.pptAlgebra Revision.ppt
Algebra Revision.ppt
 
Cbse Class 12 Maths Sample Paper 2013 Model 3
Cbse Class 12 Maths Sample Paper 2013 Model 3Cbse Class 12 Maths Sample Paper 2013 Model 3
Cbse Class 12 Maths Sample Paper 2013 Model 3
 
The solution-of-problem
The solution-of-problemThe solution-of-problem
The solution-of-problem
 
The solution of problem
The solution of problemThe solution of problem
The solution of problem
 
Sample question paper 2 with solution
Sample question paper 2 with solutionSample question paper 2 with solution
Sample question paper 2 with solution
 
1.3 solving equations t
1.3 solving equations t1.3 solving equations t
1.3 solving equations t
 
resposta do capitulo 15
resposta do capitulo 15resposta do capitulo 15
resposta do capitulo 15
 
Basic algebra for entrepreneurs
Basic algebra for entrepreneurs Basic algebra for entrepreneurs
Basic algebra for entrepreneurs
 
Basic algebra for entrepreneurs
Basic algebra for entrepreneurs Basic algebra for entrepreneurs
Basic algebra for entrepreneurs
 

More from Anantha Bellary (20)

A study on ratio analysis at vst tillers tractors final
A study on ratio analysis at vst tillers tractors finalA study on ratio analysis at vst tillers tractors final
A study on ratio analysis at vst tillers tractors final
 
An Organisational study at TUMUL, Tumkur, Karnataka
An Organisational study at TUMUL, Tumkur, KarnatakaAn Organisational study at TUMUL, Tumkur, Karnataka
An Organisational study at TUMUL, Tumkur, Karnataka
 
Fii
FiiFii
Fii
 
Certificates
CertificatesCertificates
Certificates
 
Annexure
AnnexureAnnexure
Annexure
 
Body
BodyBody
Body
 
Chapter titles
Chapter titlesChapter titles
Chapter titles
 
Bibliography
BibliographyBibliography
Bibliography
 
Abstract
AbstractAbstract
Abstract
 
Table of content
Table of contentTable of content
Table of content
 
21 chap
21 chap21 chap
21 chap
 
20 chap
20 chap20 chap
20 chap
 
19 chap
19 chap19 chap
19 chap
 
18 chap
18 chap18 chap
18 chap
 
17 chap
17 chap17 chap
17 chap
 
16 chap
16 chap16 chap
16 chap
 
14 chap
14 chap14 chap
14 chap
 
13 chap
13 chap13 chap
13 chap
 
12 chap
12 chap12 chap
12 chap
 
11 chap
11 chap11 chap
11 chap
 

9 chap

  • 1. Free GK Alerts- JOIN OnlineGK to 9870807070 9. SURDS AND INDICES IMPORTANT FACTS AND FORMULA K 1. LAWS OF INDICES: (i) am x an = am + n eG (ii) am / an = am-n (iii) (am)n = amn (iv) (ab)n = anbn (v) ( a/ b )n = ( an / bn ) (vi) a0 = 1 2. SURDS: Let a be a rational number and n be a positive integer such that a1/n = nsqrt(a) 3. LAWS OF SURDS: (i) n√a = a1/2 in is irrational. Then nsqrt(a) is called a surd of order n. nl (ii) n √ab = n √a * n √b (iii) n √a/b = n √a / n √b (iv) (n √a)n = a (v) m√(n√(a)) = mn√(a) (vi) (n√a)m = n√am eO I SOLVED EXAMPLES Ex. 1. Simplify : (i) (27)2/3 (ii) (1024)-4/5 (iii)( 8 / 125 )-4/3 Sol . (i) (27)2/3 = (33)2/3 = 3( 3 * ( 2/ 3)) = 32 = 9 Th (ii) (1024)-4/5 = (45)-4/5 = 4 { 5 * ( (-4) / 5 )} = 4-4 = 1 / 44 = 1 / 256 (iii) ( 8 / 125 )-4/3 = {(2/5)3}-4/3 = (2/5){ 3 * ( -4/3)} = ( 2 / 5 )-4 = ( 5 / 2 )4 = 54 / 24 = 625 / 16 Ex. 2. Evaluate: (i) (.00032)3/5 (ii)l (256)0.16 x (16)0.18. Sol. (i) (0.00032)3/5 = ( 32 / 100000 )3/5. = (25 / 105)3/5 = {( 2 / 10 )5}3/5 = ( 1 / 5 )(5 * 3 / 5) = (1/5)3 = 1 / 125 (ii) (256)0. 16 * (16)0. 18 = {(16)2}0. 16 * (16)0. 18 = (16)(2 * 0. 16) * (16)0. 18
  • 2. =(16)0.32 * (16)0.18 = (16)(0.32+0.18) = (16)0.5 = (16)1/2 = 4. 196 Ex. 3. What is the quotient when (x-1 - 1) is divided by (x - 1) ? K Sol. x-1 -1 = (1/x)-1 = _1 -x * 1 = -1 x-1 x-1 x (x - 1) x Hence, the required quotient is -1/x Ex. 4. If 2x - 1 + 2x + 1 = 1280, then find the value of x. eG Sol. 2x - 1 + 2X+ 1 = 1280  2x-1 (1 +22) = 1280 2x-1 = 1280 / 5 = 256 = 28  x -1 = 8  x = 9. Hence, x = 9. in Ex. 5. Find the value of [ 5 ( 81/3 + 271/3)3]1/ 4 nl Sol. [ 5 ( 81/3 + 271/3)3]1/ 4 = [ 5 { (23)1/3 + (33)1/3}3]1/ 4 = [ 5 { (23 * 1/3)1/3 + (33 *1/3 )1/3}3]1/ 4 = {5(2+3)3}1/4 = (5 * 53)1/ 4 =5(4 * 1/ 4) = 51 = 5. eO Ex. 6. Find the Value of {(16)3/2 + (16)-3/2} Sol. [(16)3/2 +(16)-3/2 = (42)3/2 +(42)-3/2 = 4(2 * 3/2) + 4{ 2* (-3/2)} = 43 + 4-3 = 43 + (1/43) = ( 64 + ( 1/64)) = 4097/64. Ex. 7. If (1/5)3y = 0.008, then find the value of(0.25)y. Th Sol. (1/5)3y = 0.008 = 8/1000 = 1/125 = (1/5)3  3y = 3  Y = 1.  (0.25)y = (0.25)1 = 0.25.
  • 3. Ex. 8. Find the value of (243)n/5  32n + 1 9n  3n -1 . Sol. (243)n/5 x32n+l = 3 (5 * n/5)  32n+l _ = 3n 32n+1 (32)n  3n - 1 32n  3n - 1 32n  3n-l K = 3n + (2n + 1) = 3(3n+1) = 3(3n+l)-(3n-l) = 32 = 9. 32n+n-1 3(3n-1) Ex. 9. Find the value Of (21/4-1)(23/4+21/2+21/4+1) eG Sol. Putting 21/4 = x, we get : (21/4-1) (23/4+21/2+21/4+1)=(x-1)(x3+x2+x+1) , where x = 21/4 =(x-1)[x2(x+1)+(x+1)] =(x-1)(x+1)(x2+1) = (x2-1)(x2+1) Ex. 10. Find the value of 62/3  3√67 3 √66 in =(x4-1) = [(21/4)4-1] = [2(1/4*4) –1] = (2-1) = 1. nl Sol. 62/3  3√67 = 62/3  (67)1/3 = 62/3  6(7 * 1/3) = 62/3  6(7/3) 3 √66 (66)1/3 6(6 * 1/3) 62 =62/3  6((7/3)-2) = 62/3  61/3 = 61 = 6. Ex. 11. If x= y , y=z and z=xc,then find the value of abc. a b eO Sol. z1= xc =(ya)c [since x= ya] =y(ac) = (zb)ac [since y=zb] =zb(ac)= zabc  abc = 1. Ex. 12. Simplify [(xa / xb)^(a2+b2+ab)] * [(xb / xc )^ b2+c2+bc)] * [(xc/xa)^(c2+a2+ca)] Th Sol. Given Expression = [{x(o - b)}^(a2 + b2 + ob)].['(x(b - c)}^ (b2 + c2 + bc)].['(x(c - a)}^(c2 + a2 + ca]) = [x(a - b)(a2 + b2 + ab) . x(b - c) (b2 +c2+ bc).x(c- a) (c2 + a2 + ca)] = [x^(a3-b3)].[x^(b3-e3)].[x^(c3-a3)] = x^(a3-b3+b3-c3+c3-a3) = x0 = 1.
  • 4. Ex. 13. Which is larger √2 or 3√3 ? K Sol. Given surds are of order 2 and 3. Their L.C.M. is 6. Changing each to a surd of order 6, we get: √2 = 21/2 = 2((1/2)*(3/2)) =23/6 = 81/6 = 6√8 3 √3= 31/3 = 3((1/3)*(2/2)) = 32/6 = (32)1/6 = (9)1/6 = 6√9. eG Clearly, √9 > √8 and hence 3√3 > √2. 6 6 Ex. 14. Find the largest from among 4√6, √2 and 3√4. Sol. Given surds are of order 4, 2 and 3 respectively. Their L.C,M, is 12, Changing each to a surd of order 12, we get: 4 √6 = 61/4 = 6((1/4)*(3/3)) = 63/12 = (63)1/12 = (216)1/12. 3 in √2 = 21/2 = 2((1/2)*(6/6)) = 26/12 = (26)1/12 = (64)1/12. √4 = 41/3 = 4((1/3)*(4/4)) = 44/12 = (44)1/12 = (256)1/12. Clearly, (256)1/12 > (216)1/12 > (64)1/12 nl Largest one is (256)1/12. i.e. 3√4 . eO Th
  • 5. . Th eO nl in eG K