Irrational Equation 8

It is obvious that $x^{2} \neq 1$ . Therefore, by moving $\sqrt[3]{1- x^{2} }$ to the left hand side, the given equation can be written in the form: $\sqrt[3]{ \frac{1+x}{1-x} } + 15 \sqrt[3]{\frac{1-x}{1+x}} = 8$ . The substitution $y = \sqrt[3]{ \frac{1+x}{1-x} }$ yields $y + \frac{15}{y} = 8$ i.e. $y^{2} - 8y +15 = 0$ , and its solutions are 3 and 5.

Now, we have two cases:

(i) $y = 3 \Rightarrow \sqrt[3]{ \frac{1+x}{1-x} } =3 \Rightarrow 1+x = 27(1-x) \Rightarrow x_{1} = \frac{13}{14}$

(ii) $y = 5 \Rightarrow \sqrt[3]{ \frac{1+x}{1-x} } =5 \Rightarrow 1+x = 125(1-x) \Rightarrow x_{2} = \frac{62}{63}$

Hence, we get the answer 1.913.