Simplify: [(-2/3)⁻²]³ × (1/3)⁻⁴ × 3⁻¹ × 1/6

Solution:

Given, the expression is [(-2/3)⁻²]³ × (1/3)⁻⁴ × 3⁻¹ × 1/6

We have to simplify the given expression.

(a^m)ⁿ = a^mn

So, [(-2/3)⁻²]³ = (-2/3)^{-2 × 3}

= (-2/3)⁻⁶

Using law of exponents,

a⁻ⁿ = 1/aⁿ

So, (-2/3)⁻⁶ = 1/(-2/3)⁻⁶

= (-3/2)⁶

Similarly, (1/3)⁻⁴ = 1/(1/3)⁻⁴

= 3⁴

3⁻¹ = 1/3

We know, (-a)ⁿ = aⁿ, when n is even number

So, (-3/2)⁶ = (3/2)⁶

Using law of exponents,

a^m/b^m = (a/b)^m

(3/2)⁶ = 3⁶/2⁶

Now, [(-2/3)⁻²]³ × (1/3)⁻⁴ × 3⁻¹ × 1/6 = 3⁶/2⁶ × 3⁴ × 1/3 × 1/6

= 3⁶/2⁶ × 3³ × 1/(3 × 2)

= 3⁶/2⁶ × 3² × 1/2

Using law of exponents,

a^m × aⁿ = a^{m + n}

= 3^{6 + 2}/2^{6 + 1}

= 3⁸/2⁷

Therefore, the required value is 3⁸/2⁷.

✦ Try This: Simplify: [(3/4)⁻²]³ × (1/4)⁻⁴ × 4⁻¹ × 1/16

☛ Also Check: NCERT Solutions for Class 8 Maths Chapter 12

NCERT Exemplar Class 8 Maths Chapter 8 Problem 98(ii)

Summary:

On simplifying [(-2/3)⁻²]³ × (1/3)⁻⁴ × 3⁻¹ × 1/6, we get 3⁸/2⁷ using the law of exponents (a^m)ⁿ = a^mn

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