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B. xy²

Let a = x³y² = x × x × x × y × y

And b = xy³ = x × y × y × y

⇒ HCF of a and b = HCF (x³y², xy³) = x × y × y = xy²

[Since, HCF is the product of the smallest power of each common prime factor involved in the numbers]

Beep duration of first device = 60 seconds 

Beep duration of second device = 62 seconds 

∴ Interval of beeping together = LCM (60, 62) 

Prime factorization: 

60 = 2² × 3 × 5 

62 = 2 × 31 

∴ LCM = 2² × 3 × 5 × 31 = 1860 seconds = 1860/60 = 31min 

Hence, they will beep together again at 10 : 31 a.m

Answer is (A) irrational number

Given:

Beep duration of first device = 60 seconds

Beep duration of second device = 62 seconds

To find the interval of beeping together, we need to find the LCM of 60 and 62 first.

60 = 2 x 2 × 3 × 5

62 = 2 × 31

LCM = 2 x 2 × 3 × 5 × 31 = 1860 seconds

Convert time into mins:

160 sec = 1860/60 mins = 31 min

Electronic devices will beep after every 31 minutes.

Hence, both devices will beep together again at 10:31 a.m. again.

p/q (q ≠ 0) is a rational number.

and p/q is terminating decimal.

Prime factor of q will be in the form 2^m x 5ⁿ 

where m, n is non-negative integers.

Answer is (c) 3

43/(2² x 5³)

Multiply by 2 in numerator and denominator of number

43/(2² x 5³) = 86/10³ = 0.086

Let x = 2 √3 – 1 be a rational number. 

x = 2 √3 – 1 

⇒ x² = (2 √3 – 1)² 

⇒ x² = (2 √3 )² + (1)² – 2(2 √3)(1) 

⇒ x² = 12 + 1 - 4 √3

⇒ x² – 13 = - 4 √3

⇒ \(\frac{13-x^2}4\) = √3

Since x is rational number, x² is also a rational number. 

⇒ 13 - x² is a rational number

⇒ \(\frac{13-x^2}4\) is a rational number

⇒ √3 is a rational number 

But √3 is an irrational number, which is a contradiction. 

Hence, our assumption is wrong. 

Thus, (2 √3 – 1) is an irrational number.

Let \(\frac{1}{\sqrt3}\) be rational.

∴ \(\frac{1}{\sqrt3}\) = \(\frac{a}{b}\), where a, b are positive integers having no common factor other than 1

∴ \(\sqrt3\) = \(\frac{b}{a}\).....(1)

Since a, b are non-zero integers, \(\frac{b}{a}\) is rational.

Thus, equation (1) shows that \(\sqrt3\) is rational.

This contradicts the fact that \(\sqrt3\) is rational.

The contradiction arises by assuming \(\sqrt3\) is rational.

Hence, \(\frac{1}{\sqrt3}\) is irrational.

(i) True.

(ii) True.

(iii) False.

Reason:

Sum of two irrational can be rational. Let us take an example,

Consider two irrational numbers: (3 + √2) and (3 – √2)

Sum: (3 + √2) + (3 – √2) = 3 + √2 + 3 – √2 = 6 which is rational.

(iv) False.

Reason:

Product of two irrational numbers can be rational. Let us take an example,

Consider two irrational numbers: (3 + √2) and (3 – √2)

Product: (3 + √2)(3 – √2 ) 

= (3)² – (√2 )² 

= 9 – 2 = 7; which is rational

(v) True.

(vi) True.

(i) The sum of two rationals is always rational - True 

(ii) The product of two rationals is always rational - True 

(iii) The sum of two irrationals is an irrational - False 

Counter example: 2 + √3 and 2 - √3 are two irrational numbers. But their sum is 4, which is a rational number. 

(iv) The product of two irrationals is an irrational – False 

Counter example: 

2 √3 and 4 √3 are two irrational numbers. But their product is 24, which is a rational number. 

(v) The sum of a rational and an irrational is irrational - True 

(vi) The product of a rational and an irrational is irrational - True

(i) The sum of two rationals is always rational - True 

(ii) The product of two rationals is always rational - True 

(iii) The sum of two irrationals is an irrational - False Counter example: 2 + √3 and 2 - √3 are two irrational numbers. But their sum is 4, which is a rational number. 

(iv) The product of two irrationals is an irrational – False Counter example: 2 √3 and 4 √3 are two irrational numbers. But their product is 24, which is a rational number. 

(v) The sum of a rational and an irrational is irrational - True 

(vi) The product of a rational and an irrational is irrational - True

Correct option is (C) \(\frac3{11}\)

(b) \(\frac{19}{80}\)

\(\frac{19}{80}\) = \(\frac{19}{2^4\times5}\)

We know 2 and 5 are not factors of 19, so it is in its simplest form. 

And (2⁴ × 5) = (2^m × 5ⁿ) 

Hence, \(\frac{19}{80}\) is a terminating decimal.

Correct option is (B) 10

Required number is HCF of 420 and 130.

\(\therefore\) Required number = HCF (420, 130)

= 10

Correct option is (B) 10

(i) \(\frac{23}{8}\)

We have, \(\frac{23}{8}\) and here the denominator is 8. 

⇒ 8 = 2³ x 5 

We see that the denominator 8 of \(\frac{23}{8}\) is of the form 2^m x 5ⁿ, where m, n are non-negative integers. 

Hence, \(\frac{23}{8}\) has terminating decimal expansion. And, the decimal expansion of \(\frac{23}{8}\) terminates after three places of decimal.

(ii) \(\frac{125}{441}\)

We have, \(\frac{125}{441}\) and here the denominator is 441. 

⇒ 441 = 3² x 7² 

We see that the denominator 441 of \(\frac{125}{441}\) is not of the form 2^m x 5ⁿ, where m, n are non-negative integers. 

Hence, \(\frac{125}{441}\) has non-terminating repeating decimal expansion.

(iii) \(\frac{35}{50}\)

We have, \(\frac{35}{50}\) and here the denominator is 50. 

⇒ 50 = 2 x 5² 

We see that the denominator 50 of \(\frac{35}{50}\) is of the form 2^m x 5ⁿ, where m, n are non-negative integers. 

Hence, \(\frac{35}{50}\) has terminating decimal expansion. And, the decimal expansion of \(\frac{35}{50}\) terminates after two places of decimal.

(iv) \(\frac{77}{210}\)

We have, \(\frac{77}{210}\) and here the denominator is 210. 

⇒ 210 = 2 x 3 x 5 x 7 

We see that the denominator 210 of \(\frac{77}{210}\) is not of the form 2^m x 5ⁿ, where m, n are non-negative integers. 

Hence, \(\frac{77}{210}\) has non-terminating repeating decimal expansion.

(v) \(\frac{129}{(2^2 \times 5^7 \times 7^{17})}\)

We have, \(\frac{129}{(2^2 \times 5^7 \times 7^{17})}\) and here the denominator is 2² x 5⁷ x 7¹⁷. 

Clearly, 

We see that the denominator is not of the form 2^m x 5ⁿ, where m, n are non-negative integers. 

And hence, \(\frac{129}{(2^2 \times 5^7 \times 7^{17})}\) has non-terminating repeating decimal expansion.

(vi) \(\frac{987}{10500}\)

We have, \(\frac{987}{10500}\) 

But, \(\frac{987}{10500}\) = \(\frac{47}{500}\) (reduced form) 

And now the denominator is 500. 

⇒ 500 = 2² x 5³ 

We see that the denominator 500 of \(\frac{47}{500}\) is of the form 2^m x 5ⁿ, where m, n are non-negative integers. 

Hence, \(\frac{987}{10500}\) has terminating decimal expansion. And, the decimal expansion of \(\frac{987}{10500}\) terminates after three places of decimal.

13/3125=13/2⁰x 5⁵

This, shows that the prime factorisation of the denominator is of the form 2^m × 5ⁿ.

Hence, it has terminating decimal expansion.

Given number = 6ⁿ = (2 × 3)ⁿ 

The prime factors here are 2 and 3 only. 

To be end with 0; 6ⁿ should have a prime factor 5 and also 2. So, 6ⁿ can’t end with zero.

(i) Required number of minutes is the LCM of 18 and 12.

18 = 2 x 3²

and 12 =  2² x 3 

.'. LCM of 18 and 12 = 2² x 3² = 36

Hence, Ravish and Priya will meet again at the starting point after 36 minutes.

(ii) LCM of numbers

(iii) Healthy competition is necessary for personal development and progress

Correct option is (A) 36 minutes

Correct option is (C) Non-terminating and repeating

\(\frac{9}{17};\) denominator = 17

\(\therefore\) Prime factors of denominator is other than 2 or 5.

\(\therefore\) \(\frac{9}{17}\) has non-terminating and repeating decimals.

Correct option is (C) Non-terminating and repeating

Solution:

They will be meet again after LCM of both values at the starting point.
To get the LCM we have to factorize the number.
18        = 2 × 3 × 3
12        = 2 × 2 × 3
LCM    = 2 × 2 × 3 × 3 = 36 
Therefore, they will meet together at the starting point after 36 minutes.

If 4ⁿ ends with 0, then it must have 5 as a factor. 

But we know the only prime factor of 4ⁿ is 2. 

Also we know from the fundamental theorem of arithmetic that prime factorization of each number is unique. 

Hence, 4ⁿ can never end with the digit 0.

17/30 = 17/ (2 × 3 × 5) 

We know that 2,3 and 5 are not the factors of 17. 

So, 17/30 is in its simplest form. 

Also, 30 = 2 × 3 × 5 ≠ (2^m × 5ⁿ ) 

Hence, 17/30 is a non-terminating decimal.

(d) 81 

Let the two numbers be x and y. 

It is given that: x = 54 

HCF = 27 

LCM = 162 

We know, 

x × y = HCF × LCM 

⇒ 54 × y = 27 × 162 

⇒ 54 y = 4374 

⇒ ∴ y = 4374/54 

= 81

(d) 81 

Let the two numbers be x and y. 

It is given that: 

x = 54 

HCF = 27 

LCM = 162 

We know, 

x × y = HCF × LCM 

⇒ 54 × y = 27 × 162 

⇒ 54 y = 4374

⇒ ∴ y = \(\frac{4374}{54}\) = 81

HCF of two numbers = 18 

Product of two numbers = 12960 

Let their LCM be x. 

Using the formula, product of two numbers = HCF × LCM 

we conclude that 

12960 = 18 × x 

x = 12960/18 = 720 

Hence, their LCM is 720.

Let the two numbers be x and y 

It is given that: 

x = 81 

HCF = 27 and LCM = 162 

We know, Product of two numbers = HCF × LCM 

⇒ x × y = 27 × 162 

⇒ 81 × y = 4374 

⇒ y = 4374/81 = 54 

Hence, the other number is y is 54.

Let the two numbers be a and b. 

Let the value of a be 161. 

Given: HCF = 23 and LCM = 1449 

We know, a × b = HCF × LCM 

⇒ 161 × b = 23 × 1449 

⇒ ∴ b = 23 ×1449/ 161 = 33327/161 = 207 

Hence, the other number b is 207.

Here, the integer ‘n’ is of the form 5q+1.

⇒ n= 5q+1 

On squaring it, 

⇒ n²= (5q+1)² 

⇒ n²= (25q²+10q+1) 

⇒ n²= 5(5q²+2q)+1 

⇒ n²= 5m+1, where m is some integer. [For m = 5q²+2q] 

Therefore, the square of any positive integer of the form 5q + 1 is of the same form.

Let n = 5q + 1 where q is a positive integer
∴ n² = (5q + 1)²
= 25q² + 10q + 1
= 5(5q² + 2q) + 1
= 5m + 1, where m is some integer
Hence, the square of any positive integer of the form 5q + 1 is of the same form.

From Euclid’s division lemma, 

a = bq+r ; where 0 < r < b 

Putting b=4 for the question, 

⇒ a = 4q + r, 0 < r < 4 

For r = 0, we get a = 4q, which is an even number. 

For r = 1, we get a = 4q + 1, which is an odd number. 

On squaring, 

⇒ a² = (4q + 1)² 

= 16q² + 1 + 8q 

= 8(2q² + q) + 1 

= 8m + 1, where m = 2q² + q 

For r = 2, we get a = 4q + 2 = 2(2q + 1), which is an even number. 

For r = 3, we get a = 4q + 3, which is an odd number. 

On squaring, 

⇒ a² = (4q + 3)² 

= 16q² + 9 + 24q 

= 8(2q² + 3q + 1) + 1

= 8m + 1, where m = 2q² + 3q + 1 

Thus, the square of an odd integer is of the form 8q + 1, for some integer q.

Let any positive integer ‘n’ be of the form 3q or, 3q+1 or 3q+2. (From Euclid’s division lemma for b= 3) 

If n= 3q, 

Then, on squaring 

⇒ n²= (3q)² = 9q² 

⇒ n²= 3(3q²) 

⇒ n²= 3m, where m is some integer [m = 3q²] 

If n= 3q+1, 

Then, on squaring 

⇒ n²= (3q+1)² = 9q² + 6q + 1 

⇒ n²= 3(3q² +2q) + 1 

⇒ n²= 3m + 1, where m is some integer [m = 3q² +2q] 

If n= 3q+2, 

Then, on squaring 

⇒ n²= (3q+2)² = 9q² + 12q + 4 

⇒ n²= 3(3q² + 4q + 1) + 1 

⇒ n²= 3m, where m is some integer [m = 3q² + 4q + 1] 

Thus, it is observed that the square of any positive integer is of the form 3m or 3m + 1 but not of the form 3m + 2.

Let ‘a’ be any positive integer. 

Then,

According to Euclid’s division lemma, 

a = bq+r 

According to the question, when b = 5. 

a = 5k + r, 0 < r < 5 

When r = 0, we get, a = 5k 

a² = 25k² = 5(5k²) = 5q, where q = 5k² 

When r = 1, we get, a = 5k + 1 

a² = (5k + 1)² 

= 25k² + 1 + 10k 

= 5k(5k + 2) + 1 

= 5q + 1, where q = k(5k + 2) 

When r = 2, we get, a = 5k + 2 

a² = (5k + 2)² 

= 25k² + 4 + 20k 

= 5(5k² + 4k) + 4 

= 4q + 4, where q = 5k² + 4k 

When r = 3, we get, a = 5k + 3 

a² = (5k + 3)² 

= 25k² + 9 + 30k 

= 5(5k² + 6k + 1) + 4 

= 5q + 4, where q = 5k² + 6k + 1 

When r = 4, we get, a = 5k + 4 

a² = (5k + 4)² 

= 25k² + 16 + 40k 

= 5(5k² + 8k + 3) + 1 

= 5q + 1, where q = 5k² + 8k + 3 

Therefore, the square of any positive integer is of the form 5q or, 5q + 1 or 5q + 4 for some integer q.

By Euclid’s division Algorithm
a = bm + r, where 0 ≤ r ≤ b
Put b = 4
a = 4m + r, where 0 ≤ r ≤ 4
If r = 0, then a = 4m
If r = 1, then a = 4m + 1
If r = 2, then a = 4m + 2
If r = 3, then a = 4m + 3
Now, (4m)² = 16m²
= 4 × 4m²
= 4q where q is some integer
(4m + 1)² = (4m)² + 2(4m)(1) + (1)²
= 16m² + 8m + 1
= 4(4m² + 2m) + 1
= 4q + 1 where q is some integer
(4m + 2)² = (4m)² + 2(4m)(2)+(2)²
= 16m² + 24m + 9
= 16m² + 24m + 8 + 1
= 4(4m² + 6m + 2) + 1
= 4q + 1, where q is some integer

Hence, the square of any positive integer is of the form 4q or 4q + 1 for some integer m​

Since any positive integer n is of the form 2p or, 2p + 1

When n = 2p,

then n² = 4p² = 4a

where a = p²

When n = 2p + 1,

then n² = (2p + 1)² = 4p² + 4p + 1

⇒ 4p(p + 1) + 1 ⇒ 4m + 1

where m = p(p + 1)

Therefore square of any positive integer is of the form 4q or 4q + 1 for some integer q

Let us find the HCF of each pair of numbers. 

(i) 396 = 231 × 1 + 165 

231 = 165 × 1 + 66 

165 = 66 × 2 + 33 

66 = 33 × 2 + 0 

Therefore, HCF = 33. Hence, numbers are not co-prime. 

(ii) 2160 = 847 × 2 + 466 

847 = 466 × 1 + 381 

466 = 381 × 1 + 85 

381 = 85 × 4 + 41 

85 = 41 × 2 + 3 

41 = 3 × 13 + 2 

3 = 2 × 1 + 1 

2 = 1 × 2 + 0 

Therefore, the HCF = 1. Hence, the numbers are co-prime.

According to Euclid’s division lemma,

a=bq+r

According to the question,

When b = 4.

a = 4k + r, 0 < r < 4

When r = 0, we get, a = 4k

a² = 16k² = 4(4k²) = 4q, where q = 4k²

When r = 1, we get, a = 4k + 1

a² = (4k + 1)² = 16k² + 1 + 8k = 4(4k + 2) + 1 = 4q + 1, where q = k(4k + 2)

When r = 2, we get, a = 4k + 2

a² = (4k + 2)² = 16k² + 4 + 16k = 4(4k² + 4k + 1) = 4q, where q = 4k² + 4k + 1

When r = 3, we get, a = 4k + 3

a² = (4k + 3)² = 16k² + 9 + 24k = 4(4k² + 6k + 2) + 1

= 4q + 1, where q = 4k² + 6k + 2

Therefore, the square of any positive integer is either of the form 4q or 4q + 1 for some integer q.

Hence Proved.

Correct option is (C) Z or l

Z or I = {...., -3, -2, -1, 0, 1, 2, 3, .....}

Correct option is  C) Z or l

Any positive odd integer is of the form 2q + 1, where q is a whole number. 

Therefore, (2q + 1)² = 4q² + 4q + 1 = 4q (q + 1) + 1, ...... (1) 

q (q + 1) is either 0 or even. So, it is 2m, where m is a whole number. 

Therefore, (2q + 1)2 = 4.2 m + 1 = 8 m + 1. [From (1)]

Correct option is (A) 3

Given that \(p^3+q^3=r^3\)

\(\Rightarrow\) \(p^3=r^3-q^3\) \(=(r-q)(r^2+qr+q^2)\)

\(\Rightarrow\) \(log_p\,p^3\) \(=log_p((r-q)(r^2+qr+q^2))\)   (By taking \(log_p\) both sides)

\(\Rightarrow\) \(log_p(r-q)+log_p(r^2+qr+q^2)\) \(=3\,log_p\,p=3\)

Correct option is (A) 3

Correct option is (C) 27 abc

We have \(\sqrt[3]a+\sqrt[3]b+ \sqrt[3]c=0\)

\(\Rightarrow\) \(a^\frac13+b^\frac13+c^\frac13=0\)

\(\Rightarrow\) \(a^\frac13+b^\frac13=-c^\frac13\)    ___________(1)

\(\Rightarrow\) \((a^\frac13+b^\frac13)^3=(-c^\frac13)^3\)  (By cubing both sides)

\(\Rightarrow(a^\frac13)^3+(b^\frac13)^3+3a^\frac13b^\frac13(a^\frac13+b^\frac13)=-(c^\frac13)^3\)  \((\because(a+b)^3=a^3+b^3+3ab(a+b)\,\&\,(ab)^3=a^3b^3)\)

\(\Rightarrow a+b+3\,a^\frac13\,b^\frac13\times-c^\frac13=-c\)   (From (1))

\(\Rightarrow\) a+b+c \(=3\,a^\frac13\,b^\frac13c^\frac13\)

\(\Rightarrow\) \((a+b+c)^3\) \(=(3\,a^\frac13\,b^\frac13c^\frac13)^3\)   (By cubing both sides)

\(\Rightarrow\) \((a+b+c)^3=27\,abc\)    \((\because(ab)^m=a^mb^m)\)

Correct option is (C) 27 abc

Correct option is (A) 0

\(\because\) First whole number is zero.

\(\therefore\) Product of first 2019 whole numbers is zero.

\((\because\) 0 . 1 . 2 . 3 . 4 ....... 2019 = 0)

Correct option is  A) 0

Correct option is (B) 2

\(x^2=({\sqrt7-\sqrt3})({\sqrt7+\sqrt3})\)

= 7 - 3 = 4 = \(2^2\)

\(\Rightarrow\) x = 2

Correct option is B) 2

Correct option is (C) 2⁷ × 3⁴ × 7 × 11

LCM of the numbers \(2^7\times3^4\times7\) and \(2^3\times3^4\times11\) is  \(2^7\times3^4\times7\times11.\)

Correct option is C) 2⁷ × 3⁴ × 7 × 11 

Correct option is (A) \(\frac{13}{4}\)

\(3.25=\frac{325}{100}\)

\(=\frac{325\div25}{100\div25}=\frac{13}{4}\)

Correct option is  \(\frac{13}{4}\)

Correct option is (B) -2/3

\((\frac{-2}{3})^\frac27\times(\frac{-2}{3})^\frac57\) \(=(\frac{-2}{3})^{\frac27+\frac57}\)

\(=(\frac{-2}{3})^1=\frac{-2}{3}\)

Correct option is  B) -2/3

Correct option is (D) -6/8

\(\frac{-6}{8}=\frac{-6\div2}{8\div2}\) \(=\frac{-3}{4}\)

Hence, \(\frac{-6}{8}\) is equivalent to rational number \(\frac{-3}{4}.\)

Correct option is  D) -6/8

Correct option is (C) \(\sqrt[3]4+1\)

A rationalising factor of a given irrational number is an irrational number whose multiple with given irrational number gives a rational number.

(A) \((16^\frac13-4^\frac13+1)(4^\frac13+2)\)

\(=64^\frac13-16^\frac13+4^\frac13+2.16^\frac13-2.4^\frac13+2\)

\(=(4^3)^\frac13+16^\frac13-4^\frac13+2\)

\(=4+2+16^\frac13-4^\frac13\)

\(=6+16^\frac13-4^\frac13\) which is an irrational number.

Thus, \((4^\frac13+2)\) is not a rationalising factor of \((16^\frac13-4^\frac13+1).\)

(B) \((16^\frac13-4^\frac13+1)(4^\frac13-2)\)

\(=64^\frac13-16^\frac13+4^\frac13-2.16^\frac13+2.4^\frac13-2\)

\(=(4^3)^\frac13-3.16^\frac13+3.4^\frac13-2\)

\(=4-2-3.16^\frac13+3.4^\frac13\)

\(=2-3.16^\frac13+3.4^\frac13\) which is an irrational number.

Thus, \((4^\frac13-2)\) is not a rationalising factor of \((16^\frac13-4^\frac13+1).\)

(C) \((16^\frac13-4^\frac13+1)(4^\frac13+1)\)

\(=64^\frac13-16^\frac13+4^\frac13+16^\frac13-4^\frac13+1\)

\(=(4^3)^\frac13+1\)

= 4+1 = 5 which is a rational number.

Thus, \((4^\frac13+1)\) is a rationalising factor of \((16^\frac13-4^\frac13+1).\)

(D) \((16^\frac13-4^\frac13+1)(4^\frac13-1)\)

\(=64^\frac13-16^\frac13+4^\frac13-16^\frac13+4^\frac13-1\)

\(=(4^3)^\frac13-2.16^\frac13+2.4^\frac13-1\)

\(=4-1-2.16^\frac13+2.4^\frac13\)

\(=3-2.16^\frac13+2.4^\frac13\) which is an irrational number.

Thus, \((4^\frac13-1)\) is not a rationalising factor of \((16^\frac13-4^\frac13+1).\)

Correct option is (C) \(^3\sqrt{4}\) + 1 

Correct option is (B) 5^1/3 + 5^1/3

\(\because\left((25)^\frac13+\frac1{(25)^\frac13}-1\right)\left(5^\frac13+\frac1{5^\frac13}\right)\)

\(=\left(5^\frac23+\frac1{5^\frac23}-1\right)\left(5^\frac13+\frac1{5^\frac13}\right)\)

\(=5^\frac23.5^\frac13+\cfrac{5^\frac13}{5^\frac23}-5^\frac13+\cfrac{5^\frac23}{5^\frac13}+\frac1{5^\frac23\,5^\frac13}-\frac1{5^\frac13}\)

\(=5^\frac33+\frac1{5^\frac13}-5^\frac13+5^\frac13+\frac1{5^\frac33}-\frac1{5^\frac13}\)   \((\because a^m.a^n=a^{m+n}\,and\,\cfrac{a^m}{a^n}=a^{m-n})\)

\(=5+\frac15=\frac{26}5\) which is a rational number.

Thus, \((5^{\frac{1}{3}}+\frac1{5^{\frac{1}{3}}})\) is a rationalising factor of \((25^\frac13+\frac1{25^\frac13}-1).\)

Correct option is (B) 5^1/3 + 5^1/3

Correct option is (A) \(\sqrt[5]{3^1}\)

\(3^\frac{1}{5}=\sqrt[5]{3}\)

= \(\sqrt[5]{3^1}\)

Correct option is  A) \(\sqrt[5]{3^1}\)