Question

Use a suitable identity to find the product of (3a - 1/3) (3a + 1/3)​

Answer 1

$(3a+\frac{1}{3})(3a-\frac{1}{3})=\frac{81a^2-1}{9}$

Step-by-step explanation:

Given : Expression $(3a+\frac{1}{3})(3a-\frac{1}{3})$

To find : Use a suitable identity to find the product of expression ?

Solution :

Expression $(3a+\frac{1}{3})(3a-\frac{1}{3})$

Using algebraic identity,

$(x+y)(x-y)=x^2-y^2$

Here, x=3a and $y=\frac{1}{3}$

Substitute the values,

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

$(3a+\frac{1}{3})(3a-\frac{1}{3})=9a^2-\frac{1}{9}$

$(3a+\frac{1}{3})(3a-\frac{1}{3})=\frac{81a^2-1}{9}$

Therefore, $(3a+\frac{1}{3})(3a-\frac{1}{3})=\frac{81a^2-1}{9}$

#Learn more

Using suitable identities evaluate (1÷3a-b)^3

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Answer 2

Step-by-step explanation:

Expression (3a+\frac{1}{3})(3a-\frac{1}{3})(3a+

3

1

)(3a−

3

1

)

To find : Use a suitable identity to find the product of expression ?

Solution :

Expression (3a+\frac{1}{3})(3a-\frac{1}{3})(3a+

3

1

)(3a−

3

1

)

Using algebraic identity,

(x+y)(x-y)=x^2-y^2(x+y)(x−y)=x

2

−y

2

Here, x=3a and y=\frac{1}{3}y=

3

1

Substitute the values,

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

3

1

)(3a−

3

1

)=(3a)

2

−(

3

1

)

2

(3a+\frac{1}{3})(3a-\frac{1}{3})=9a^2-\frac{1}{9}(3a+

3

1

)(3a−

3

1

)=9a

2

−

9

1

(3a+\frac{1}{3})(3a-\frac{1}{3})=\frac{81a^2-1}{9}(3a+

3

1

)(3a−

3

1

)=

9

81a

2

−1

Therefore, (3a+\frac{1}{3})(3a-\frac{1}{3})=\frac{81a^2-1}{9}(3a+

3

1

)(3a−

3

1

)=

9

81a

2

−1

#Learn more

Using suitable identities evaluate (1÷3a-b)^3

https://brainly.in/question/3925260