VAlue of x

You realize that (7x-2) and (7x+5) differ by 7. What two whole cube numbers differ by 7? 1 and 8, of course. Then you solve 7x-2=1 for x=(3/7). Translate it into decimals and that's your answer. Often, it helps working backwards to solve problems that seem complex.

Let $\space \begin{cases} 7x-2 = (a-b)^3 & ...(1) \\ 7x+5 = (a+b)^3 &...(2) \end{cases} \\ \Rightarrow \sqrt [3] {7x-2} + \sqrt [3] {7x+5} = 3 \quad \Rightarrow a-b+a+b = 3 \quad \Rightarrow a = \frac{3}{2}$

Eq . 2 $-$ Eq. 1:

$(a+b)^3-(a-b)^3 = 7\quad \Rightarrow 6a^2b + 2b^3 = 7\quad \Rightarrow 6\left(\frac{9}{4}\right)b + 2b^3 = 7\\ \Rightarrow 4b^3+27b-14=0 \quad \Rightarrow (2b-1)(2b^2+b+14) = 0 \quad \Rightarrow b = \frac{1}{2}$

From Eq. 1: $\space 7x-2 = (a-b)^3 = \left(\frac{3}{2}-\frac{1}{2}\right)^3 = 1\quad \Rightarrow x = \boxed{\frac{3}{7}}$

Let $7x - 2 = a$

So , given equation becomes -

$\sqrt[3]{a} + \sqrt[3]{a + 7} = 3$

As $R.H.S = 3$ is a positive integer , for $L.H.S$ to be a positive integer ,

Let us consider $a , a + 7$ as perfect positive cubes. $\Rightarrow a = m^3 , a + 7 = n^3$ , for some positive $m,n$

So , $m^3 - n^3 = 7$

$(m - n)(m^2 + mn + n^2) = 7$

$Case - 1:$

If $m - n = 7$ , $m^2 + mn + n^2 = 1$ , then , $m > 7$ , So , $m^2 + mn + n^2$ becomes too large to be equal to 1. Hence , this case is not possible.

$Case - 2:$

If $m - n = 1$ , $m^2 + mn + n^2 = 7$ , then ,

$m = 1 + n$ .... Substituting this in $m^2 + mn + n^2 = 7$ ,

$(1 + n)^2 + (1 + n)n + n^2 = 7$

$1 + 2n + n^2 + n + n^2 + n^2 = 7$

$3n^2 + 3n + 1 = 7$

$3n^2 + 3n - 6 = 0$

$(3n - 3)(n + 2) = 0$

$3n - 3 = 0 \Rightarrow 3n = 3 \Rightarrow n = 1$

$n + 2 = 0 \Rightarrow n = -2$

But, n is positive , hence , $\boxed{n = 1}$

So , $m = n + 1 \Rightarrow m = 1 + 1 \Rightarrow \boxed{m = 2}$

Hence , $a = 1^3 = 1$ , $a + 7 = 2^3 = 8$

So , $7x - 2 = 1$

$7x = 3$

$x = \frac{3}{7}$

$\huge \boxed{x = 0.428}$

(7x-2)^(1/3) + (7x+5)^(1/3) = 3

((7x-2)^(1/3))^3 + ((7x+5)^(1/3))^3 = (3)^3

(7x-2) + (7x+5) = 9

14x + 3 = 9

14x = 6

x = 6/14