ដោះស្រាយ sqrt{a^2-70a+25}=sqrt[3]{x^9}

$$\sqrt{ { a }^{ 2 } -70a+25 } = \sqrt[3]{ { x }^{ 9 } }$$

ដោះស្រាយសម្រាប់ a

$$a=-\sqrt{x^{6}+1200}+35$$

$$a=\sqrt{x^{6}+1200}+35\text{, }x\geq 0$$

ដោះស្រាយសម្រាប់ x

$$\left\{\begin{matrix}x=\sqrt[6]{a^{2}-70a+25}\text{, }&a\geq 20\sqrt{3}+35\text{ or }a\leq 35-20\sqrt{3}\\x=\frac{2^{\frac{2}{3}}\sqrt[3]{-\sqrt{a^{2}-70a+25}+\sqrt{-3a^{2}+210a-75}}}{2}\text{; }x=\frac{2^{\frac{2}{3}}\sqrt[3]{-\sqrt{a^{2}-70a+25}-\sqrt{-3a^{2}+210a-75}}}{2}\text{, }&a=35-20\sqrt{3}\text{ or }a=20\sqrt{3}+35\end{matrix}\right.$$