Question

If 2x - 1/2x = 4, then find - 8x 3 -1/8x 3

Answer 1

2x – 1/2x = 4
Cubing on both sides
(2x – 1/2x)³ = 4³
8x³ – 1/8x³ –3×2x×1/2x×(2x–1/2x) = 64
8x³ – 1/8x³ –3×4 = 64
8x³–1/8x³ = 64+12
8x³ – 1/8x³ = 76

Answer 2

Given:

$\mathbf{2x-\dfrac{1}{2x}=4}$        ...1)

To Find:

$\mathbf{Find \ value\ of\ 8x^{3}-\dfrac{1}{8x^{3}}=?}$

Solution:

We know the identity equation:

$\mathbf{(A-B)^{3}=A^{3}-B^{3} -3A\times B(A-B)}$     ...2)

Taking cube of both side of equation 1):

$\mathbf{\left ( 2x-\dfrac{1}{2x} \right )^{3}=4^{3}}$

L.H.S of above equation can be solved by with help of equation 2):

$\mathbf{(2x)^{3}-\left ( \dfrac{1}{2x} \right )^{3} -3\times 2x\times \dfrac{1}{2x}(2x-\dfrac{1}{2x})=64}$

$\mathbf{8x^{3}-\dfrac{1}{8x^{3}} -3\times (2x-\dfrac{1}{2x})=64}$        ...3)

With help of equation 1):

$\mathbf{8x^{3}-\dfrac{1}{8x^{3}} -3\times 4=64}$

$\mathbf{8x^{3}-\dfrac{1}{8x^{3}} -12=64}$

$\mathbf{8x^{3}-\dfrac{1}{8x^{3}} =76}$

This is answer.Means

$\mathbf{Value\ of\ 8x^{3}-\dfrac{1}{8x^{3}} \ is\ 76}$