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Question : 46 of 86

Marks: +1, -0

If

a = 3^{\frac{1}{4}} + 3^{-\frac{1}{4}}

b = 3^{\frac{1}{4}} - 3^{-\frac{1}{4}}

then the value of

(a^{2}+b^{2})^{2}

is :

67
65
$\frac{64}{3}$
None

Solution:

Concept:

Use algebraic expansion and simplification to find the expression value.

Explanation:

Given

a = 3^{1/4} + 3^{-1/4}

and

b = 3^{1/4} - 3^{-1/4}

.

Compute

a^2 = (3^{1/4} + 3^{-1/4})^2 = 3^{1/2} + 2\cdot 3^{1/4}\cdot 3^{-1/4} + 3^{-1/2} = 3^{1/2} + 2 + 3^{-1/2}

.

Similarly,

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

.

Add:

a^2 + b^2 = (3^{1/2} + 2 + 3^{-1/2}) + (3^{1/2} - 2 + 3^{-1/2}) = 2\cdot 3^{1/2} + 2\cdot 3^{-1/2} = 2(3^{1/2} + 3^{-1/2})

.

Now square:

(a^2+b^2)^2 = 4(3^{1/2} + 3^{-1/2})^2

.

Compute

(3^{1/2} + 3^{-1/2})^2 = 3 + 2 + 3^{-1} = 5 + \frac{1}{3} = \frac{16}{3}

.

Thus

(a^2+b^2)^2 = 4 \cdot \frac{16}{3} = \frac{64}{3}

.

Answer:

\frac{64}{3}

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