Đáp án:
a) $S=\left\{\dfrac{\sqrt[3]{4}+1}{2}\right\}$
b) $S=\left\{-\dfrac{1}{1+\sqrt[3]{2}}\right\}$
Giải thích các bước giải:
a) $8x^3-12x^2+6x-5=0$
$⇔(2x)^3-3.(2x)^2.1+3.2x.1^2-1^3-4=0$
$⇔(2x-1)^3=4$
$⇔2x-1=\sqrt[3]{4}$
$⇔x=\dfrac{\sqrt[3]{4}+1}{2}$
Vậy $S=\left\{\dfrac{\sqrt[3]{4}+1}{2}\right\}$
b) $3x^3+3x^2+3x=-1$
$⇔(x^2+3x^2+3x+1)+2x^3=0$
$⇔(x+1)^3=-2x^3$
$⇔x+1=-\sqrt[3]{2}x$
$⇔x=-\dfrac{1}{1+\sqrt[3]{2}}$
Vậy $S=\left\{-\dfrac{1}{1+\sqrt[3]{2}}\right\}$.