Question

решите 21 плииизпрошуу

Answer 1

$\tt36(x-8)=(x-8)^3\\6^2\cdot(x-8)-(x-8)^3=0\\(x-8)(6^2-(x-8)^2)=0\\(x-8)(6-(x-8))(6+x-8)=0\\(x-8)(6-x+8)(x-2)=0\\(x-8)(14-x)(x-2)=0\\\\ x-8=0; \ \ \ \ 14-x=0; \ \ \ \ x-2=0\\x_1=8; \ \ \ \ \ \ \ \ \ x_2=14; \ \ \ \ \ \ \ x_3=2$