Question

x+ 1/2x=4 , x^3+1/8x^3​

Answer 1

Solution:

Given That:

$\tt \longrightarrow x + \dfrac{1}{2x} = 4$

Cubing both sides, we get:

$\tt \longrightarrow \bigg(x + \dfrac{1}{2x} \bigg)^{3} = {4}^{3}$

We know:

$\bigstar \: \: \underline{ \boxed{ \tt {(x + y)}^{3} = {x}^{3} + 3 {x}^{2}y +3x {y}^{2} + {y}^{3} }}$

Using this identity, we get:

$\tt \longrightarrow {x}^{3} + 3 \cdot {x}^{2} \cdot \dfrac{1}{2x} + 3 \cdot x \cdot \bigg( \dfrac{1}{2x} \bigg)^{2} + \bigg(\dfrac{1}{2x} \bigg)^{3} =64$

$\tt \longrightarrow {x}^{3} + \dfrac{3x}{2} +\dfrac{3}{4x} + \bigg(\dfrac{1}{2x} \bigg)^{3} =64$

$\tt \longrightarrow {x}^{3} + \dfrac{1}{8 {x}^{3} } + \dfrac{3x}{2} +\dfrac{3}{4x} =64$

$\tt \longrightarrow {x}^{3} + \dfrac{1}{8 {x}^{3} } + \dfrac{3}{2} \bigg(x+\dfrac{1}{2x} \bigg)=64$

Substituting the value from above, we get:

$\tt \longrightarrow {x}^{3} + \dfrac{1}{8 {x}^{3} } + \dfrac{3}{2} \times 4=64$

$\tt \longrightarrow {x}^{3} + \dfrac{1}{8 {x}^{3} } + 6=64$

$\tt \longrightarrow {x}^{3} + \dfrac{1}{8 {x}^{3} } =58$

Which is our required answer.

Learn More:

Algebraic Identities.

• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²
• (a + b)² + (a - b)² = 2(a² + b²)
• (a + b)² - (a - b)² = 4ab
• a² - b² = (a + b)(a - b)
• (a + b)³ = a³ + 3ab(a + b) + b³
• (a - b)³ = a³ - 3ab(a - b) - b³
• a³ + b³ = (a + b)(a² - ab + b²)
• a³ - b³ = (a - b)(a² + ab + b²)
• (x + a)(x + b) = x² + (a + b)x + ab
• (x + a)(x - b) = x² + (a - b)x - ab
• (x - a)(x + b) = x² - (a - b)x - ab
• (x - a)(x - b) = x² - (a + b)x + ab
• (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac