Question

simplify (27)6/5_(27)1/5​

Answer 1

$27 \frac{6}{5} - 27 \frac{1}{5} \\ \\ 27( \frac{6}{5} - \frac{1}{5} ) \\ \\ 27( \frac{6 - 1}{5} ) \\ \\ 27 \frac{5}{5} = 27 \times 1 = 27$

Answer 2

Given:

$(27)^{\dfrac{6}{5} }- (27)^{\dfrac{1}{5} }$

We have to find, the value of $(27)^{\dfrac{6}{5} }- (27)^{\dfrac{1}{5} }$ is:

Solution:

∴ $(27)^{\dfrac{6}{5} }- (27)^{\dfrac{1}{5} }$

27 = 3 × 3 × 3 = $3^{2}$

$=(3^3)^{\dfrac{6}{5} }- (3^3)^{\dfrac{1}{5} }$

Using the exponential identity:

$(a^m)^{n}$ = $a^{mn}$

$=(3)^{3\times \dfrac{6}{5} }- (3)^{3\times \dfrac{1}{5} }$

Taking common as $3^{\dfrac{3}{5} }$, we get

= $3^{\dfrac{3}{5} }$$(3^6-1)$

= $3^{\dfrac{3}{5} }$$(729-1)$

= 728· $3^{\dfrac{3}{5} }$

∴ $(27)^{\dfrac{6}{5} }- (27)^{\dfrac{1}{5} }$ = $3^{\dfrac{3}{5} }$$(3^6-1)$ or, 728· $3^{\dfrac{3}{5} }$

Thus, the value of $(27)^{\dfrac{6}{5} }- (27)^{\dfrac{1}{5} }$ is equal to "$3^{\dfrac{3}{5} }$$(3^6-1)$ or, 728· $3^{\dfrac{3}{5} }$"