Question

Simplify : [(64)^-1/6 × (216)^-1/3 × (81)^1/4]/[(512)^-1/3 × (16)^1/4 × (9)^-1/2

Answer 1

$\dfrac{(64)^{-\frac{1}{6}}\times (216)^{-\frac{1}{3}}\times (81)^{\frac{1}{4}}}{(512)^{-\frac{1}{3}}\times (16)^{\frac{1}{4}}\times (9)^{-\frac{1}{2}}}=3$

Step-by-step explanation:

$\dfrac{(64)^{-\frac{1}{6}}\times (216)^{-\frac{1}{3}}\times (81)^{\frac{1}{4}}}{(512)^{-\frac{1}{3}}\times (16)^{\frac{1}{4}}\times (9)^{-\frac{1}{2}}}$

Step 1. Express the square terms of both the numerator and the denominator as powers of their factors as required.

$=\dfrac{(2^{6})^{-\frac{1}{6}}\times (6^{3})^{-\frac{1}{3}}\times (3^{4})^{\frac{1}{4}}}{(8^{3})^{-\frac{1}{3}}\times (2^{4})^{\frac{1}{4}}\times (3^{2})^{-\frac{1}{2}}}$

Step 2. Use the formula $(a^{m})^{n}=a^{mn}$ and simplify the terms.

$=\dfrac{2^{-\frac{6}{6}}\times 6^{-\frac{3}{3}}\times 3^{\frac{4}{4}}}{8^{-\frac{3}{3}}\times 2^{\frac{4}{4}}\times 3^{-\frac{2}{2}}}$

$=\dfrac{2^{-1}\times 6^{-1}\times 3^{1}}{8^{-1}\times 2^{1}\times 3^{-1}}$

Step 3. Express the terms of both the numerator and the denominator as proruct and power of their prime factors as required.

$=\dfrac{2^{-1}\times (2\times 3)^{-1}\times 3^{1}}{(2^{3})^{-1}\times 2^{1}\times 3^{-1}}$

Step 4. Use the formula $(a\times b)^{m}=a^{m}\times b^{m}$ and use the formula used in Step 2.

$=\dfrac{2^{-1}\times 2^{-1}\times 3^{-1}\times 3^{1}}{2^{-3}\times 2^{1}\times 3^{-1}}$

$=\dfrac{2^{-1-1}\times 3^{-1+1}}{2^{-3+1}\times 3^{-1}}$

$=\dfrac{2^{-2}\times 3^{0}}{2^{-2}\times 3^{-1}}$

Step 5. Use the formula $\dfrac{a^{m}}{a^{n}}=a^{m-n}$ to simplify.

$=2^{-2-(2)}\times 3^{0-(-1)}$

$=2^{-2+2}\times 3^{0+1}$

$=2^{0}\times 3^{1}$

$=1\times 3$

• Since $2^{0}=1$ and $3^{1}=3$

$=\bold{3}$

Answer 2

Step-by-step explanation:

\dfrac{(64)^{-\frac{1}{6}}\times (216)^{-\frac{1}{3}}\times (81)^{\frac{1}{4}}}{(512)^{-\frac{1}{3}}\times (16)^{\frac{1}{4}}\times (9)^{-\frac{1}{2}}}=3

(512)

−

3

1

×(16)

4

1

×(9)

−

2

1

(64)

−

6

1

×(216)

−

3

1

×(81)

4

1

=3

Step-by-step explanation:

\dfrac{(64)^{-\frac{1}{6}}\times (216)^{-\frac{1}{3}}\times (81)^{\frac{1}{4}}}{(512)^{-\frac{1}{3}}\times (16)^{\frac{1}{4}}\times (9)^{-\frac{1}{2}}}

(512)

−

3

1

×(16)

4

1

×(9)

−

2

1

(64)

−

6

1

×(216)

−

3

1

×(81)

4

1

Step 1. Express the square terms of both the numerator and the denominator as powers of their factors as required.

=\dfrac{(2^{6})^{-\frac{1}{6}}\times (6^{3})^{-\frac{1}{3}}\times (3^{4})^{\frac{1}{4}}}{(8^{3})^{-\frac{1}{3}}\times (2^{4})^{\frac{1}{4}}\times (3^{2})^{-\frac{1}{2}}}=

(8

3

)

−

3

1

×(2

4

)

4

1

×(3

2

)

−

2

1

(2

6

)

−

6

1

×(6

3

)

−

3

1

×(3

4

)

4

1

Step 2. Use the formula (a^{m})^{n}=a^{mn}(a

m

)

n

=a

mn

and simplify the terms.

=\dfrac{2^{-\frac{6}{6}}\times 6^{-\frac{3}{3}}\times 3^{\frac{4}{4}}}{8^{-\frac{3}{3}}\times 2^{\frac{4}{4}}\times 3^{-\frac{2}{2}}}=

8

−

3

3

×2

4

4

×3

−

2

2

2

−

6

6

×6

−

3

3

×3

4

4

=\dfrac{2^{-1}\times 6^{-1}\times 3^{1}}{8^{-1}\times 2^{1}\times 3^{-1}}=

8

−1

×2

1

×3

−1

2

−1

×6

−1

×3

1

Step 3. Express the terms of both the numerator and the denominator as proruct and power of their prime factors as required.

=\dfrac{2^{-1}\times (2\times 3)^{-1}\times 3^{1}}{(2^{3})^{-1}\times 2^{1}\times 3^{-1}}=

(2

3

)

−1

×2

1

×3

−1

2

−1

×(2×3)

−1

×3

1

Step 4. Use the formula (a\times b)^{m}=a^{m}\times b^{m}(a×b)

m

=a

m

×b

m

and use the formula used in Step 2.

=\dfrac{2^{-1}\times 2^{-1}\times 3^{-1}\times 3^{1}}{2^{-3}\times 2^{1}\times 3^{-1}}=

2

−3

×2

1

×3

−1

2

−1

×2

−1

×3

−1

×3

1

=\dfrac{2^{-1-1}\times 3^{-1+1}}{2^{-3+1}\times 3^{-1}}=

2

−3+1

×3

−1

2

−1−1

×3

−1+1

=\dfrac{2^{-2}\times 3^{0}}{2^{-2}\times 3^{-1}}=

2

−2

×3

−1

2

−2

×3

0

Step 5. Use the formula \dfrac{a^{m}}{a^{n}}=a^{m-n}

a

n

a

m

=a

m−n

to simplify.

=2^{-2-(2)}\times 3^{0-(-1)}=2

−2−(2)

×3

0−(−1)

=2^{-2+2}\times 3^{0+1}=2

−2+2

×3

0+1

=2^{0}\times 3^{1}=2

0

×3

1

=1\times 3=1×3

Since 2^{0}=12

0

=1 and 3^{1}=33

1

=3

=\bold{3}=3