Question

If `25x^2 + 1/(4x^2) = 20,` find the value of `125x^3 + 1/(8x^3)`

Answer 1

Given: `25x^2 + 1/(4x^2) = 20,`

Let’s express it in terms of a binomial

Here, let a = 5x, b = `1/(2x),`

Then, a² = 25x², b² = `1/(4x^2),`

∴ a² + b² = 20

Finding a + b

(a + b)² = a² + b² + 2ab

(a + b)² = 20 + 2`(5x)(1/(2x))` 

(a + b)² = 20 + 2`(5/2)`

(a + b)² = 20 + 5

(a + b)² = 25

∴ a + b = `+-sqrt25` 

∴ a + b = ±5

Using the identity,

a³ + b³ = (a + b)³ − 3ab(a + b)

a³ + b³ = (5)³ − 3`(5x)(1/(2x))(`5)

a³ + b³ = 125 − 3`(5/2)(5)`

a³ + b³ = 125 − `(15/2)(5)`

a³ + b³ = 125 − `(75/2)`

a³ + b³ = `(125-75)/2`

∴ a³ + b³ = `175/2`

∴ a³ + b³ = ±87.5

Hence, `125x^3 + 1/(8x^3) = a^3 + b^3 = +-87.5`