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Algebra and Applications Test 2

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Question : 4 of 40

Marks: +1, -0

If

x+\frac{1}{2x}=2

, find the value of

8x^3+\frac{1}{x^3}.

48
88
40
44

Solution: 👈: Video Solution

Concept:

Use the identity

(a+b)^3=a^3+b^3+3ab(a+b)

.

Explanation:

Given:

x+\frac{1}{2x}=2

.

Multiply both sides by

2

:

2x+\frac{1}{x}=4

.

Cube both sides:

\left(2x+\frac{1}{x}\right)^3=4^3

.

Expand:

(2x)^3+\left(\frac{1}{x}\right)^3+3(2x)\left(\frac{1}{x}\right)\left(2x+\frac{1}{x}\right)=64

.

Simplify:

8x^3+\frac{1}{x^3}+6\left(2x+\frac{1}{x}\right)=64

.

Substitute

2x+\frac{1}{x}=4

:

8x^3+\frac{1}{x^3}+6(4)=64

.

So,

8x^3+\frac{1}{x^3}+24=64

.

Therefore,

8x^3+\frac{1}{x^3}=40

.

Answer:

The value is

40

, so the correct option is C.

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