91Ó°ÊÓ

The given equation is $x^{\frac{2}{3}} + 9x^{\frac{1}{3}} + 8 = 0$.

Simplifying $x^{\frac{2}{3}} + 9x^{\frac{1}{3}} + 8 = 0$.

${\left(x^{\frac{1}{3}}\right)}^2 + 9x^{\frac{1}{3}} + 8 = 0$

Now, Substituting $u=x^{\frac{1}{3}}$, we get:-

$u^2+9u+8=0$

Using Quadratic formula :-

$u^2+9u+8=0$

$$\begin{gathered} u_{1, 2}=\frac{-9Â\pm \sqrt{9^2-4· 1· 8}}{2· 1} \\ {u}_{1, 2}=\frac{-9Â\pm  7}{2· 1} \\ {u}_1=\frac{-9+7}{2· 1}, u_2=\frac{-9-7}{2· 1} \\ u=-1, u=-8 \end{gathered}$$

The value of u are -1 and -8 , Using these value of u to solve for x , we get:-

$$\begin{gathered} -1=x^{\frac{1}{3}} \\ x=-1 \\ -8=x^{\frac{1}{3}} \\ x=-512 \end{gathered}$$

Therefore , The value of x are -1 and -512 .

Substitute the value of $x=-1$in equation $x^{\frac{2}{3}} + 9x^{\frac{1}{3}} + 8 = 0$, we get:-

$\begin{array}{rcl}{\left(-1\right)}^{\frac{2}{3}} + 9{\left(-1\right)}^{\frac{1}{3}} + 8 & =&  1-9+8\\ 1-9+8& =& 0\end{array}$

Similarly , Substitute the value of $x=-512$in equation $x^{\frac{2}{3}} + 9x^{\frac{1}{3}} + 8 = 0$, we get:-

$\begin{array}{rcl}{\left(-512\right)}^{\frac{2}{3}} + 9{\left(-512\right)}^{\frac{1}{3}} + 8 & =&  64-72+8\\ 64-72+8& =& 0\end{array}$

Since, for the value of $x=-1, x=-512$ the equation satisfied the condition. Hence, The solution of the equation is $x=-1, x=-512$