Question

Find the value of `8^(2/3) + root(3)(27^2) - 32^((-3)/5)`

Answer 1

Given,

`8^(2/3) + root(3)(27^2) - 32^((-3)/5)`

We need to simplify the given terms.

Thus, `8^(2/3) + root(3)(27^2) - 32^((-3)/5)`

⇒ `(2^3)^(2/3) + root(3)((3^3)^2) - (2^5)^((-3)/5)`  ...`[∴ root(n)(a) = a^(1/n), "when"  a ≠ 0]`

⇒ `(2)^(3 xx 2/3) + (3)^(6 xx 1/3) - 2^(5 xx (-3)/5)`  ...[∴ (a^m)ⁿ = a^mn)]

⇒ `(2)^2 + (3)^2 - 2^-3`

⇒ `4 + 9 - 1/8`   ...`[∴ a^-n = 1/a^n]`

⇒ `(13 - 1/8) = (104 - 1)/8`

= `103/8`

= `12 7/8`

Hence, `8^(2/3) + root(3)(27^2) - 32^((-3)/5) = 12 7/8`.