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Exponents with like Bases

If x =21/3 ...

Question

If $x = 2^{1 / 3} - 2^{- (1 / 3)}$ , then $2 x^{3} + 6 x$ is equal to

A

0

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B

3

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C

3/2

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D

6

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Solution

The correct option is B 3
$x^{3} = [2^{1 / 2} - 2^{- (1 / 3)}]^{3} = [2^{(1 / 3)}]^{3} - [2^{- (1 / 3)}]^{3} - 3 (2^{1 / 3}) [2^{- (1 / 3)}] [2^{1 / 3} - 2^{- (1 / 3)}]$
$= 2 - \frac{1}{2} - 3 x$
or $x^{3} + 3 x = \frac{4 - 1}{2} = \frac{3}{2}$
or $2 x^{2} + 6 x = 3$

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Similar questions

x = 2^{\frac{1}{3}} - 2^{\frac{- 1}{3}}

2 x^{3} + 6 x

Q. Choose the correct answer from the alternatives given.
If

x = 2^{\frac{1}{3}} + 2^{\frac{- 1}{3}}

then the value of

2 x^{3} - 6 x

x = 2^{1 / 3} + 2^{- 1 / 3}

2 x^{3} - 6 x - 5 = 0

x = 2 + 2^{1 / 3} + 2^{2 / 3}

x^{3} - 6 x^{2} + 6 x =

x = \sqrt[3]{2} - \sqrt[3]{\frac{1}{2}}

2 x^{3} + 6 x =

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