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Question : 47 of 86

Marks: +1, -0

If

a=\sqrt[3]{\sqrt{2}+1}-\sqrt[3]{\sqrt{2}-1}

, then the value of

a^{3}+3a-2

is

3
0
2
1

Solution:

Concept:

Express

a

as

x-y

and use the identity

a^3 = (x-y)^3

to relate

a^3

and

a

.

Explanation:

Let

x = \sqrt[3]{\sqrt{2}+1}

and

y = \sqrt[3]{\sqrt{2}-1}

, so

a = x - y

.

Then

x^3 = \sqrt{2}+1

and

y^3 = \sqrt{2}-1

.

Cube

a

:

a^3 = (x-y)^3 = x^3 - y^3 - 3xy(x-y) = (\sqrt{2}+1) - (\sqrt{2}-1) - 3xy \cdot a = 2 - 3xy a

.

Now compute

xy = \sqrt[3]{(\sqrt{2}+1)(\sqrt{2}-1)} = \sqrt[3]{2-1} = \sqrt[3]{1} = 1

.

Substitute

xy = 1

:

a^3 = 2 - 3a

, so

a^3 + 3a = 2

.

Thus

a^3 + 3a - 2 = 0

.

Answer:

0

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