Question

If $x+\frac{1}{2x}=3$, find the value of $8x^3+\frac{1}{x^3}$.

• A. 186
• B. 180
• C. 164
• D. 160

Answer 1

Answer:

B. 180

Solution:

Given:

$x+\frac{1}{2x}=3$

Formula used:

If $ax+b(\frac{1}{x})=k$

On cubing both side we get

(ax)³ + (b/x)³ = k³ - 3kab

Calculation:

Multiply $x+\frac{1}{2x}=3,$ by 2 on both side

⇒$2(x + \frac{1}{2x}) = 3\times{2}$

⇒ $2x + \frac{1}{x} = 6$

On cubing both side we get

⇒ $(2x + \frac{1}{x})^3 = 6^3$

⇒ $8x^3+\frac{1}{x^3}=216-3\times{6}\times{2}$

⇒ $8x^3+\frac{1}{x^3}=216-36$

⇒$8x^3+\frac{1}{x^3}=180$

∴ The value of $8x^3+\frac{1}{x^3}$ is 180.