Question

3x в 6 +6x в 4 - x во 2 -2=0

Answer 1

$3x^6+6x^4-x^2-2=0\\\\3x^4(x^2+2)-(x^2+2)=0\\\\(x^2+2)(3x^4-1)=0\\\\x^2+2\ne 0\; ,\; \; (x^2+2>0)\; \\\\3x^4-1=0\; \; \to \; \; (\sqrt3x^2-1)(\sqrt3x^2+1)=0\\\\\sqrt3x^2+1\ne 0\; \; ,\; \; (\sqrt3x^2+1>0)\; \; ,\\\\\sqrt3x^2-1=0\; \; \to \; \; (\sqrt[4]3x-1)(\sqrt[4]3x+1)=0\\\\x_1=\frac{1}{\sqrt[4]3}\; ,\; \; x_2=-\frac{1}{\sqrt[4]3}$