Question

evaluate [(64^1/2)^1/6]^2​

Answer 1

Answer:

(((64)^1/2)1/6)2

((64)^1/2*1/6*2)

((64)^1/3)1/2

(4)^1/2

2

Answer 2

Answer:

Required value is 2.

Step-by-step explanation:

Given,

${{64}^{ \frac{1}{2}^{ \frac{1}{6} } } }^{2}$

Now by prime factorisation,

$64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = {2}^{6}$

So,

${{64}^{ \frac{1}{2}^{ \frac{1}{6} } } }^{2} \\ = {( {{2}^{6}) }^{ \frac{1}{2}^{ \frac{1}{6} } } }^{2} \\ = { { {2}^{(6 \times \frac{1}{2}) } }^{ \frac{1}{6} } }^{2} \\ = { { {2}^{3} }^{ \frac{1}{6} } }^{2} \\ = { {2}^{(3 \times \frac{1}{6} )} }^{2} \\ = { {2}^{ \frac{1}{2} } }^{2} \\ = {2}^{2 \times \frac{1}{2} } \\ = {2}^{1} \\ = 2$

Here applied formulas are

${x}^{ {y}^{z} } = {x}^{y \times z} \\ {x}^{1} = x$

This is a problem of Algebra.

Some important Algebra formulas.

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab − b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)³ − 3ab(a + b)

a³ - b³ = (a -b)³ + 3ab(a - b)

a² − b² = (a + b)(a − b)

a² + b² = (a + b)² − 2ab

a² + b² = (a − b)² + 2ab

a³ − b³ = (a − b)(a² + ab + b²)

a³ + b³ = (a + b)(a² − ab + b²)

Know more about Algebra,

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2) https://brainly.in/question/1169549