How do you simplify $(27 ^(1/12) * 27 ^(-5/12))^-2$ ?

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Answer 1 · 2016-07-21T01:42:33

$9$

Keep in mind that

$(a^b)^c = a^(b*c)$

and

$a^b*a^c = a^(b+c)$

and

$a^(b/c)=root(c)(a^b)$

and

$root(b)(a^b) = a$

`

Use the first concept above and apply it to the first step.

$(27^(1/12)*27^(-5/12))^-2$

$(27^(1/12*-2/1) * 27^(-5/12*-2/1))$

$(27^(-2/12) * 27^(10/12))$

Simplify the fractions that are serving as exponents.

$(27^(-1/6)*27^(5/6))$

Now apply the second concept mentioned above. Simplify the fractions again.

$(27^(-1/6+5/6))$

$27^(4/6)$

$27^(2/3)$

Use the third concept noted at the top.

$root(3)(27^2)$

Calculate the the value inside the radicand and rewrite.

$root(3)(729)$

Find the cube root by rewriting the radicand again. Follow the fourth concept after.

$root(3)(9^3)$

$9$

How do you simplify (27 ^(1/12) * 27 ^(-5/12))^-2(27 ^(1/12) * 27 ^(-5/12))^-2?