Question

If `3x + 1/(3x) = 7, "find"  27x^3 + 1/(27x^3)`

Answer 1

Here, let a = `3x+1/(3x)`

∴ a = 7

We need to find:

`27x^3 + 1/(27x^3) = (3x + 1/(3x))^3 - 3(3x + 1/(3x))`

Using the identity:

(a + b)³ = a³ + b³ + 3ab (a + b)

`(a)^3 = x^3 + 1/x^3 + 3x xx1/x(x + 1/x)`

`∴(a)^3 = x^3 + 1/x^3 + 3(x + 1/x)`

Now, we replace `x+1/x  "with"  3x+1/(3x)`, so it converts into

`(3x + 1/(3x))^3 = 27x^3 + 1/(27x^3) + 3(3x + 1/(3x))`

Thus, rewriting the equation as

`27x^3 + 1/(27x^3) = (3x + 1/(3x))^3 - 3(3x + 1/(3x))`

= 7³ − 3 × 7

= 343 − 21

∴ `27x^3 + 1/(27x^3) = 322`