# NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions - Exercise 1.4

### Access Exercises of Class 12 Maths Chapter 1 - Relations and Functions

Exercise 1.1 Solutions : 16 Questions (14 Short Answers, 2 MCQ)

Exercise 1.2 Solutions : 12 Questions (10 Short Answers, 2 MCQ)

Exercise 1.3 Solutions : 14 Questions (12 Short Answers, 2 MCQ)

Exercise 1.4 Solutions : 13 Questions (12 Short Answers, 1 MCQ)

Miscellaneous Exercise Solutions: 19 Questions (7 Long answers, 9 Short Answer Type, 3 MCQs)

** Exercise 1.4 :**

**1. Determine whether or not each of the definition ****of * given below gives a binary operation. In ****the event that * is not a binary operation, give ****justification for this.**

**(i) On Z ^{+}, define * by a * b = a – b**

**(ii) On Z ^{+}, define * by a * b = ab**

**(iii) On R, define * by a * b = ab ^{2}**

**(iv) On Z ^{+}, define * by a * b = |a – b|**

**(v) On Z ^{+}, define * by a * b = a**

**Sol.** **(i)** On Z^{+}, is defined by a * b = a – b

It is not a binary operation as the image of (1, 2) under * is 1 * 2 = 1 – 2 = 1 ∉ Z^{+}

**(ii)** On Z^{+}, * is defined by a * b = ab.

It is seen that for each, a, b ∉ Z^{+}, there is a unique element ab in Z^{+}.

This means that * carries each pair (a, b) to a unique element

a * b = ab in Z^{+}.

Therefore, * is binary operation.

**(iii)** On R, * is defined by a * b = ab^{2}.

It is seen that for each a, b ∈ R, there is a

unique element ab^{2} in R.

This means that * carries each pair (a, b) to a unique element a * b = ab^{2} in R. Therefore, * is binary operation.

**(iv)** On Z^{+}, * is defined by a * b = |a – b|.

It is seen that for each a, b ∈ Z+, there is a unique element |a – b| in Z^{+}.

This means that * carries each pair (a, b) to a unique element a * b = |a – b| in Z^{+}.

Therefore, * is a binary operation.

**(v)** On Z^{+}, * is defined by a * b = a.

It is seen that for each a, b ∈ Z^{+}, there is a unique element a ∈ Z^{+}.

This means that * carries each pair (a, b) to a unique element

a * b = a in Z^{+}.

Therefore, * is a binary operation.

**2. For each operation * defined below, determine ****whether * is binary, commutative or associative.**

**(i) On Z, define a * b = a – b**

**(ii) On Q, define a * b = ab + 1**

$$\textbf{(iii) On Q, define a * b =}\space\frac{\textbf{ab}}{\textbf{2}}$$

**(iv) On Z ^{+}, define a * b = 2a^{b}**

**(v) On Z ^{+}, define a * b = a^{b}**

$$\textbf{(vi) On R – }\lbrace \textbf{\normalsize– 1}\rbrace\textbf{,}\space\\\textbf{define a * b =}\frac{\textbf{a}}{\textbf{b+1}}$$

**Sol.** **(i)** On Z, * is defined by a * b = a – b

a – b ∈ Z, so the operation * is binary.

It can be observed that 1 * 2 = 1 – 2 = – 1 and

2 * 1 = 2 – 1 = 1.

Therefore, 1 * 2 ≠ 2 * 1, where 1, 2 ∈ Z

Hence, the operation * is not commutative.

Also, we have

(1 * 2) * 3 = (1 – 2) * 3 = – 1 * 3 = – 1 – 3 = – 4

1 * (2 * 3) = 1 * (2 – 3) = 1* – 1 = 1 – (– 1) = 2

Therefore, (1 * 2) * 3 ≠ 1 * (2 * 3), where

1, 2, 3 ∈ Z

Hence, the operation * is not associative.

**(ii)** On Q, * is defined by a * b = ab + 1

ab + 1 ∈ Q, so operation * is binary

It is known that

ab = ba for a, b ∈ Q

Therefore, ab + 1 = ba + 1 for a, b ∈ Q

Therefore, a * b = a * b for a, b ∈ Q

Therefore, the operation * is commutative.

It can be observed that

(1 * 2) * 3 = (1 × 2 + 1) * 3 = 3 * 3 = 3 × 3 + 1 = 10

1 * (2 * 3) = 1 * (2 × 3 + 1) = 1 * 7 = 1 × 7 + 1 = 8

Therefore, (1 * 2) * 3 ≠ 1 * (2 * 3), where 1, 2, 3 ∈ Q

Therefore, the operation * is not associative.

$$\textbf{(iii)} \text{On Q, *is defined by a b a*b =}\frac{ab}{2}.\\\frac{ab}{2}\epsilon\text{Q},\space\text{so the operation * is binary.}$$

It is known that

ab = ba for a, b ∈ Q

$$\text{Therefore,}\space\frac{ab}{2} = \frac{ba}{2}\space\text{for a, b ∈ Q} $$

Therefore, a * b = b * a for a, b ∈ Q

Therefore, the operation * is commutative.

For all a, b, c ∈ Q, we have

$$\text{(a*b)*c} =\bigg(\frac{ab}{2}\bigg)*c\\=\frac{\bigg(\frac{ab}{2}\bigg)c}{2} =\frac{abc}{4}\\ a*(b*c) = a*\bigg(\frac{bc}{2}\bigg) \\=\frac{a\bigg(\frac{bc}{2}\bigg)}{2} =\frac{abc}{4}$$

Therefore, (a * b) * c = a * (b * c)

Therefore, the operation * is associative.

**(iv)** On Z^{+}, * is defined by a * b = 2^{ab}.

2^{ab} ∈ Z^{+}, so the operation * is binary operation

It is known that

ab = ba for a, b ∈ Z^{+}

Therefore, 2^{ab} = 2^{ba} for a, b ∈ Z^{+}

Therefore, a * b = b * a for a, b ∈ Z^{+}

Therefore, the operation * is commutative.

It can be observed that

(1 * 2) * 3 = 2^{(1 × 2)} * 3 = 4 * 3 = 2^{4 × 3} = 2^{4 × 3} = 212

1 * (2 * 3) = 1 * 2^{2 × 3} = 1 * 26 = 1 * 64 = 2^{64}

Therefore, the operation * is not associative.

**(v)** On Z^{+}, * is defined by a * b = a^{b},

a^{b} ∈ Z^{+}, so the operation * is binary opertion.

It can be observed that 1 * 2 = 1^{2} = 1 and 2 * 1 = 2^{1} = 2

Therefore, 1 * 2 ≠ 2 * 1 where 1 , 2 ∈ Z^{+}

Therefore, the operation * is not commutative.

It can also be observed that

(2 * 3) * 4 = 2^{3} * 4 = 8 * 4 = 8^{4} = (2^{3})^{4} = 2^{12}

2 * (3 * 4) = 2 * 3^{4} = 2 * 81 = 2^{81}

Therefore, (2 * 3 ) * 4 ≠ 2 * (3 * 4); where 2,

3, 4 ∈ Z^{+}

Therefore, the operation * is not associative.

**(vi)** On R – {– 1}, * is defined by

$$a*b = \frac{a}{b+1}.\\\frac{a}{b+1}\epsilon \text{R}\space\text{for b} ≠ – 1$$

so that operation * is binary.

$$\text{It can be observed that}\\1 * 2 = \frac{1}{2+1}=\frac{1}{3}\space\text{and}\space\\2*1 =\frac{2}{1 + 1} =\frac{2}{2}= 1$$

Therefore, 1 * 2 ≠ 2 * 1 where 1, 2 ∈ R – {– 1} Therefore, the operation * is not commutative,

It can also be observed that

$$(1*2)*3 =\frac{1}{3}*3\\=\frac{\frac{1}{3}}{3+1} =\frac{1}{12}\\1 *(2*3) = 1*\frac{2}{3+1} = 1*\frac{2}{4}\\= 1*\frac{1}{2} =\frac{1}{\frac{1}{2}+1} =\frac{1}{\frac{3}{2}}=\frac{2}{3}$$

Therefore, (1 * 2) * 3 ≠ 1 * (2 * 3) where 1, 2, 3 ∈ R – {– 1}.

Therefore, the operation * is not associative.

**3. Consider the binary operation ^ on the set {1, 2, ****3, 4, 5} defined by a ^ b = min {a, b}. Write the ****operation table of the operation.**

**Sol.** The binary operation * on the set {1, 2, 3, 4, 5} is defined as a ^ b = min {a, b} for a, b ∈ {1, 2, 3, 4,

5}.

Thus, the operation table for the given operation ^ can be given as

^ | 1 | 2 | 2 | 4 | 5 |

1 | 1 | 1 | 1 | 1 | 1 |

2 | 1 | 2 | 2 | 2 | 2 |

3 | 1 | 2 | 3 | 3 | 3 |

4 | 1 | 2 | 3 | 4 | 4 |

5 | 1 | 2 | 3 | 4 | 5 |

**4. Consider a binary operation * on set {1, 2, 3, 4, 5} ****given by the following multiplication table:**

**(i) Compute (2 * 3) * 4 and 2 * (3 * 4)**

**(ii) Is * commutative?**

**(iii) Compute (2 * 3) * (4 * 5). **

* | 1 | 2 | 3 | 4 | 5 |

1 | 1 | 1 | 1 | 1 | 1 |

2 | 1 | 2 | 1 | 2 | 1 |

3 | 1 | 1 | 3 | 1 | 1 |

4 | 1 | 2 | 1 | 4 | 1 |

5 | 1 | 1 | 1 | 1 | 5 |

**Sol.** **(i)** We have (2 * 3) * 4 = (1) * 4 = 1

and 2 * (3 * 4) = 2 * 1 = 1

**(ii)** For every a, b ∈ {1, 2, 3, 4, 5}, we have a * b = b * a.

Therefore, the operation * is commutative.

**(iii)** We have (2 * 3) = 1 and (4 * 5) = 1

Therefore, (2 * 3) * (4 * 5) = 1 * 1 = 1

**5. Let * be the binary operation of the set {1, 2, 3, ****4, 5} defined by a * b = HCF of a and b. Is the ****operation * same as the operation * defined in ****Q. 4 above? Justify your answer.**

**Sol.** The binary operation *, on the set {1, 2, 3, 4, 5} is defined as a * b = HCF of a and b, the operation table for the operation * can be given as

* | 1 | 2 | 3 | 4 | 5 |

1 | 1 | 1 | 1 | 1 | 1 |

2 | 1 | 2 | 1 | 2 | 1 |

3 | 1 | 1 | 3 | 1 | 1 |

4 | 1 | 2 | 1 | 4 | 1 |

5 | 1 | 1 | 1 | 1 | 5 |

We observe that the operation table for the operation * and operation * in Q. 4 are the same. Thus, the operation *, is same as the operation *.

**6. Let * be the binary operation on N given by **

**a * b = LCM of a and b.**

**(i) 5 * 7, 20 * 16**

**(ii) Is * commutative?**

**(iii) Is * associative ?**

**(iv) Find the identity of * in N**

**(v) Which elements of N are invertible for the ****operation * ?**

**Sol.** The binary operation * on N is defined as a * b

= LCM of a and b.

**(i)** We have 5 * 7 = LCM of 5 and 7 = 35

and 20 * 16 = LCM of 20 and 16 = 80

**(ii)** It is known that

LCM of a and b = LCM of b and a for a, b ∈ N.

Therefore, a * b = b * a. Thus, the operation * is commutative.

**(iii)** For a, b, c ∈ N, we have

(a * b) * c = (LCM of a and b) * c = LCM of a, b and c

a * (b * c) = a * (LCM of b and c) = LCM of a, b and c

Therefore, (a * b) * c = a * (b * c). Thus, the operation * is associative.

**(iv)** It is known that

LCM of a and 1 = a = LCM of 1 and a, a ∈ N

a * 1 = a = 1 * a, a ∈ N

Thus, 1 is the identity of * in N.

**(v)** An element a in N is invertible with respect to the operation *, if there exists an element b in N such that a * b = e = b * a.

Here, e = 1. This means that

LCM of a and b = 1 = LCM of b and a.

This case is possible only when a and b are equal to 1.

Thus, 1 is the only invertible element of N with respect to the operation *.

**7. Is * defined on the set {1, 2, 3, 4, 5} by a * b ****= LCM of a and b a binary operation? Justify ****your answer.**

**Sol.** The operation * on the set A = {1, 2, 3, 4, 5} is defined as a * b = LCM of a and b. Now, 2 * 3 = LCM of 2 and 3 = 6. But 6 does not belong to the given set. Hence, the given operation * is not a binary operation.

**8. Let * be the binary operation on N defind by ****a * b = HCF of a and b. Is * commutative?****Is * associative ? Does there exist identity for ****this binary operation on N?**

**Sol.** The binary operation * on N is defined as a * b = HCF of a and b.

It is known that HCF of a and b = HCF of b and a for a, b ∈ N.

Therefore, a * b = b * a. Thus, the operation * is commutative.

For a, b, c ∈ N, we have (a * b) * c = (HCF of a and b) * c = HCF of a, b and c.

a * (b * c) = a * (HCF of b and c) = HCF of a, b and c

Therefore, (a * b) * c = a * (b * c)

Thus, the operation * is associative.

Now, an element e ∈ N will be the identity for the operation * if a * e = a = e * a, for ∀ a ∈ N. But this relation is not true for any a ∈ N.

Thus, the operation * does not have identity in N.

**9. Let * be a binary operation on the set Q of ****rational number as follows :**

**(i) a * b = a – b**

**(ii) a * b = a ^{2} + b^{2}**

**(iii) a * b = a + ab**

**(iv) a * b = (a – b) ^{2}**

$$\textbf{(v)\space}\textbf{a*b =}\frac{\textbf{ab}}{\textbf{4}}$$

**(vi) a * b = ab ^{2}**

**Find which of the binary operations are ****commutative and which are associative.**

**Sol.** (i) On Q, the operation * is defined as a * b = a – b.

It can be observed that for 2, 3, 4 ∈ Q, we have

2 * 3 = 2 – 3 = – 1 and 3 * 2 = 3 – 2 = 1

2 * 3 ≠ 3 * 2

Thus, the operation * is not commutative.

It can also be observed that

(2 * 3) * 4 = (– 1) * 4 = – 1 – 4= – 5

and 2 * (3 * 4) = 2 * (– 1) = 2 – (– 1) = 3

Thus, the operation * is not associative.

**(ii)** On Q, the operation * is defined as a * b = a^{2} + b^{2}.

For a, b ∈ Q, we have

a * b = a^{2} + b^{2} = b^{2} + a^{2} = b * a

Therefore, a * b = b * a

Thus, the operation * is commutative.

It can be observed that

(1 * 2) * 3 = (1^{2} + 2^{2}) * 3 = (1 + 4) * 4

= 5 * 4 = 5^{2} + 4^{2} = 41

1 * (2 * 3) = 1 * (2^{2} + 3^{2})

= 1 * (4 + 9) = 1 * 13

= 1^{2} + 13^{2} = 170

(1 * 2) * 3 ≠ 1 * (2 * 3) where 1, 2, 3 ∈ Q

Thus, the opetration * is not associative.

**(iii)** On Q, the operation * is defined as a * b

= a + ab

It can be observed that

1 * 2 = 1 + 1 × 2 = 1 + 2 = 3,

2 * 1 = 2 + 2 × 1 = 2 + 2 = 4

∴ 1 * 2 ≠ 2 * 1, where 1, 2 ∈ Q

Thus, the operation * is not commutative.

It can also be observed that

(1 * 2) * 3 = (1 + 1 × 2) * 3

3 * 3 = 3 + 3 × 3 = 3 + 9 = 12

1 * (2 * 3) = 1 * (2 + 2 × 3)

= 1 * 8 = 1 + 1 × 8 = 9

∴ (1 * 2) * 3 ≠ 1 * (2 * 3) where 1, 2, 3 ∈ Q

Thus, the operation * is not associative.

**(iv)** On Q, the operation * defined by a * b = (a – b)^{2} is

For a, b ∈ Q, we have

a * b = (a – b)^{2} and b * a = (b – a)^{2}

= [–(a – b)]^{2} = (a – b)^{2}

Therefore, a * b = b * a

Thus, the operation * is associative. It can

be observed that

(1 * 2) * 3 = (1 – 2)^{2} * 3 = (– 1)^{2} * 3

= 1 * 3 = (1 – 3)^{2} = (– 2)2 = 4

1 * (2 * 3) = 1 * (2 – 3)^{2} = 1 * (– 1)^{2}

= 1 * 1 = (1 – 1)^{2 }= 0

∴ (1 * 2) * 3 ≠ 1 * (2 * 3) where 1, 2, 3 ∈ Q

Thus, the operation * is not associative.

**(v)** On Q, the operation * is defined as a * b

$$=\frac{ab}{4}.\\\text{For a, b ∈ Q, we have a * b}\\=\frac{ab}{4} = \frac{ba}{4} = b*a$$

Therefore, a * b = b * a

Thus, the operation * is commutative.

For a, b, c ∈ Q, we have (a * b) * c =

$$\frac{ab}{4}*c\\=\frac{\frac{ab}{4}.c}{4} =\frac{abc}{16}\\ a*(b*c) = a*\frac{bc}{4}\\=\frac{a.\frac{bc}{4}}{4} =\frac{abc}{16}$$

Therefore, (a * b) * c = a * (b * c). Thus, the

operation * is associative.

**(vi)** On Q, the operation * is defined as a * b = ab^{2}

It can be observed that for 2, 3 ∈ Q

2 * 3 = 2.3^{2} = 18 and 3 * 2 = 3.2^{2} = 12

Hence, 2 * 3 ≠ 3 * 2

$$\text{Also,\space}\frac{1}{2} + \frac{1}{3} = \frac{1}{2}\bigg(\frac{1}{3}\bigg)^{2}\\=\frac{1}{2}.\frac{1}{9} = \frac{1}{18}\\\frac{1}{3}*\frac{1}{2} =\frac{1}{3}\bigg(\frac{1}{2}\bigg)^{2}\\=\frac{1}{3}.\frac{1}{4} = \frac{1}{12}\\\therefore\space \frac{1}{2}*\frac{1}{3} \neq \frac{1}{3}*\frac{1}{2}\\\text{where,\space}\frac{1}{2},\frac{1}{3}\epsilon \text{Q}$$

Thus, the operation * is not commutative.

It can also be observed that for 1, 2, 3 ∈ Q

(1 * 2) * 3 = (1.2^{2}) * 3 = 4 * 3 = 4.3^{2} = 36

1 * (2 * 3) = 1 * (2.3^{2}) = 1 * 18 = 1.18^{2} = 324

(1 * 2) * 3 ≠ 1 * (2 * 3 )

$$\text{Also,\space}\bigg(\frac{1}{2}*\frac{1}{3}\bigg)*\frac{1}{4}\\=\bigg[\frac{1}{2}*\bigg(\frac{1}{3}\bigg)^{2}\bigg]*\frac{1}{4}\\= \frac{1}{18}*\frac{1}{4} = \frac{1}{18}*\bigg(\frac{1}{4}\bigg)^{2}\\=\frac{1}{18×16}\\\frac{1}{2}*\bigg(\frac{1}{3}*\frac{1}{4}\bigg) = \frac{1}{2}*\bigg[\frac{1}{3}\bigg(\frac{1}{4}\bigg)^{2}\bigg]\\=\frac{1}{2}*\frac{1}{48}\\=\frac{1}{18}= \frac{1}{2}\bigg(\frac{1}{48}\bigg)^{2} =\frac{1}{4608}$$

$$\therefore\space\bigg(\frac{1}{2}*\frac{1}{3}\bigg)*\frac{1}{4}\neq\frac{1}{2}*\bigg(\frac{1}{3}*\frac{1}{4}\bigg)\\\text{where}\space\frac{1}{4},\frac{1}{4},\frac{1}{4}\epsilon\text{Q}$$

Thus, the operation * is not associative. Hence, the operations defined in parts (ii), (iv), (v) are commutative and the operation defined in part (v) is associative.

**10. Find which of the operations given above has ****identity.**

**Sol.** An element e ∈ Q will be the identity element for the operation * if

a * e = a = e * a, ∀ a ∈ Q

**(i)** a * b = a – b

If a * e = a, a ≠ 0

⇒ a – e = a, a ≠ 0 ⇒ e = 0

Also, e * a = a

⇒ e – a = a ⇒ e = 2a

∴ e = 0 = 2a, a ≠ 0

But the identity is unique. Hence, this operation has no identity.

(ii) a * b = a^{2} + b^{2}

If a * e = a, then a^{2} + e^{2} = a

For a = –2, (– 2)^{4} + e^{2} = 4 + e^{2} ≠ – 2

Hence, there is no identity element.

(iii) a * b = a + ab

If a * e = a

⇒ a + ae = a

⇒ ae = 0 ⇒ e = 0, a ≠ 0

Also if

⇒ e * a = a ⇒ e + ea = a

$$\Rarr\space e =\frac{a}{1-a}, a\neq1\\\therefore\space e =0=\frac{a}{1-a}, a\neq 0$$

But the identity is unique. Hence this operation has no identity.

**(iv)** a * b = (a – b)^{2}

If a * e = a, then (a – e)^{2} = a. A square is always positive, so for

a = – 2, (– 2 – e)^{2} ≠ – 2

Hence, there is no identity element.

**(v)** a * b = ab/4

If a * e = a, then ae/4 = a. Hence, e = 4 is the identity element.

∴ a * 4 = 4 * a = 4a/4 = a

**(vi)** a * b = ab^{2}

If a * e = a ⇒ ae^{2} = a

⇒ e^{2} = 1 ⇒ e = ± 1

But identity is unique. Hence this operation has no identity.

Therefore only part (v) has an identity element.

**11. Let A = N × N and * be the binary operation on ****A defined by (a, b) * (c, d) = (a + c, b + d). Show ****that * is commutative and associative. Find the ****identity element for * on A, if any.**

**Sol.** Given that A = N × N

and * is a binary operation on A and is defined by

(a, b) * (c, d) = (a + c, b + d)

Let (a, b), (c, d) ∈ A

Then, a, b, c, d ∈ N

we have (a, b) * (c, d) = (a + c, b + d)

and (c, d) * (a, b) = (c + a, d + b)

= (a + c, b + d)

[Addition is commutative in the set of natural numbers]

Therefore, (a, b) * (c, d) = (c, d) * (a, b)

Therefore, the operation * is commutative.

Now, let (a, b), (c, d) (e, f) ∈ A

Then, a, b, c, d, e, f ∈ N

We have ((a, b) * (c, d)) * (e, f) = (a + c, b + d) * (e, f)

= (a + c + e, b + d + f)

and (a, b) * ((c, d) * (e, f)) = (a, b) * (c + e, d + f)

= (a + c + e, b + d + f)

∴ ((a, b) * (c, d)) * (e, f) = (a, b) * ((c, c) * (e, f))

Therefore, the operation * is associative.

An element e = (e_{1}, e_{2}) ∈ A will be an identity element for the operation * if

a * e = a = e * a ∀ a = (a_{1}, a_{2}) i.e.,

(a_{1} + e_{1}, a_{2} + e_{2}) = (a_{1}, a_{2}) = (e_{1} + a_{1}, e_{2} + a_{2})

which is not true for any element in A.

Note that a + e = a for e = 0 but 0 does not belong to N.

Therefore, the operation * does not have any identity element.

**12. State whether the following statements are true ****of false. Justify.**

**(i) For an arbitrary binary operation * on a set ****N, a * a = a ∀ a ∈ N.**

**(ii) If * is a commutative binary operation on ****N, then **

**a * (b * c) = (c * b) * a**

**Sol.** **(i)** Define an operation * on N as a * b = a + b ∀ a, b ∈ N

Then, in particular, for b = a = 3, we have

3 * 3 = 3 + 3 = 6 ≠ 3

Therefore, statement (i) is false.

**(ii)** RHS = (c * b) * a = (b * c) * a

[* is commutative]

= a * (b * c)

[Again, as * is commutative]

= LHS

Therefore, a * (b * c) = (c * b) * a

Therefore, statement (ii) is true.

**13. Consider a binary operation * on N defined as ****a * b = a ^{3} + b^{3}. Choose the correct answer.**

**(A) Is * both associative and commutative ?**

**(B) Is * commutative but not associative ?**

**(C) Is * associative but not commutative ?**

**(D) Is * neither commutative nor associative ?**

**Sol.** (B) Is * commutative but not associative ?

On N, the operation * is defined as a * b = a^{3} + b^{3}.

For, a, b ∈ N, we have

a * b = a^{3} + b^{3} = b^{3} + a^{3} = b * a

[Addition is commutative in N]

Therefore, the operation * is commutative.

It can be observed that

(1 * 2) * 3 = (1^{3} + 2^{3}) * 3

= 9 * 3 = 9^{3} + 3^{3}

= 729 + 27 = 756

1 * (2 * 3) = 1 * (2^{3} + 3^{3})

= 1 * (8 + 27) = 1 * 35

= 1^{3} + 35^{3}

= 1 + (35)^{3}

= 1 + 42875 = 42876

Therefore, (1 * 2) * 3 ≠ 1 * (2 * 3 ) where 1, 2, 3, ∈ N

Therefore, the operation * is not associative.

Hence, the operation * is commutative but not associative.