English

Evaluate: [27^2/3 + 81^1/2 ⋅ 8^1/3]^−1/3 - Mathematics

Advertisements
Advertisements

Question

Evaluate:

`[27^(2/3) + 81^(1/2) * 8^(1/3)]^((-1)/3)`

Evaluate
Advertisements

Solution

Given,

`[27^(2/3) + 81^(1/2) * 8^(1/3)]^((-1)/3)`

We have to evaluate the given terms.

Thus, `[27^(2/3) + 81^(1/2) * 8^(1/3)]^((-1)/3)`

⇒ `[(3^3)^(2/3) + (9^2)^(1/2) * (2^3)^(1/3)]^((-1)/3)`

⇒ `[(3)^(3 xx 2/3) + (9)^(2 xx 1/2) xx (2)^(3 xx 1/3)]^((-1)/3)`  ...[∴ (am)n = amn]

⇒ `[(3)^2 + 9 xx 2]^((-1)/3)`

⇒ `[9 + 18]^((-1)/3)`

⇒ `27^((-1)/3`

⇒ `[3^3]^((-1)/3) = 3^(3 xx (-1)/3`  ...`[∴ a^-n = 1/a^n]`

= `3^-1`

= `1/3`  

Hence, `[27^(2/3) + 81^(1/2) * 8^(1/3)]^((-1)/3) = 1/3`

shaalaa.com
  Is there an error in this question or solution?
Chapter 6: Indices - EXERCISE 6 [Page 66]

APPEARS IN

B Nirmala Shastry Mathematics [English] Class 9 ICSE
Chapter 6 Indices
EXERCISE 6 | Q 4. (iii) | Page 66
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×