We are given the equation \(2x + { {1} \over 3x}= 5\). We need to find the value of the expression \(27x^3+{{1} \over 8x^3}\).

Understanding the Target Expression

The target expression can be rewritten using the formula for the cube of a term: \(27x^3 = (3x)^3\) \( {{1} \over 8x^3} = ({ {1} \over 2x})^3 \) So, the expression we need to find is \( (3x)^3 + ({ {1} \over 2x})^3 \).

Relating the Given Equation to the Target Expression

Let \(a = 3x\) and \(b = { {1} \over 2x}\). We need to find the value of \(a^3 + b^3\). We know the algebraic identity: \( a^3 + b^3 = (a+b)^3 - 3ab(a+b) \)

First, let's find the product \(ab\): \( ab = (3x) \times ({ {1} \over 2x}) = {3 \over 2} \)

Next, we need to find the sum \(a+b = 3x + { {1} \over 2x}\). We are given \(2x + { {1} \over 3x} = 5\). Let's see how we can find \(3x + { {1} \over 2x}\) from this.

Notice the relationship between the terms: \( 3x = {3 \over 2} \times (2x) \) \( { {1} \over 2x} = {3 \over 2} \times ({ {1} \over 3x}) \)

Let \(A = 2x\) and \(B = { {1} \over 3x}\). We are given \(A+B=5\). The terms we need are \(a = 3x = {3 \over 2} A\) and \(b = { {1} \over 2x} = {3 \over 2} B\).

Now, let's find the sum \(a+b\): \( a+b = {3 \over 2} A + {3 \over 2} B = {3 \over 2} (A+B) \)

Substitute the given value \(A+B=5\): \( a+b = {3 \over 2} \times 5 = {15 \over 2} \)

So, we have found that \( 3x + { {1} \over 2x} = {15 \over 2} \).

Calculating the Final Value

Now we have \(a+b = {15 \over 2}\) and \(ab = {3 \over 2}\). Substitute these values into the identity for \(a^3 + b^3\): \( a^3 + b^3 = (a+b)^3 - 3ab(a+b) \) \( 27x^3 + { {1} \over 8x^3} = ({15 \over 2})^3 - 3 \times ({3 \over 2}) \times ({15 \over 2}) \)

Calculate the terms: \( ({15 \over 2})^3 = {15^3 \over 2^3} = {3375 \over 8} \) \( 3 \times ({3 \over 2}) \times ({15 \over 2}) = {9 \over 2} \times {15 \over 2} = {135 \over 4} \)

Now substitute these back: \( 27x^3 + { {1} \over 8x^3} = {3375 \over 8} - {135 \over 4} \)

To subtract the fractions, find a common denominator, which is 8: \( {135 \over 4} = {135 \times 2 \over 4 \times 2} = {270 \over 8} \)

Perform the subtraction: \( 27x^3 + { {1} \over 8x^3} = {3375 \over 8} - {270 \over 8} = {3375 - 270 \over 8} = {3105 \over 8} \)

Conclusion

The value of \(27x^3+{{1} \over 8x^3}\) is \( {3105 \over 8} \).