Thu gọn biểu thức  $\sqrt[3]{{{x^3} + 3{x^2} + 3x + 1}} - \sqrt[3]{{8{x^3} + 12{x^2} + 6x + 1}}$ ta được 

$25a$

$5a$

$ - 25{a^3}$

$ - 5a$

$\dfrac{1}{{3a}}$

$\dfrac{1}{{4a}}$ 

$ - \dfrac{1}{{3a}}$

$ - \dfrac{1}{{8a}}$

$M > N$

$M < N$

$M \ge N$

$M + N = 0$

\(\sqrt[3]{a} = 2x \Leftrightarrow {a^3} = 2x\)

\(\sqrt[3]{a} = 2x \Leftrightarrow 2a = {x^3}\)

\(\sqrt[3]{a} = 2x \Leftrightarrow a = 2{x^3}\)       

\(\sqrt[3]{a} = 2x \Leftrightarrow a = 8{x^3}\)

\(a > b \Leftrightarrow \sqrt[3]{a} > \sqrt[3]{b}\)

\(a < b \Leftrightarrow \sqrt[3]{a} < \sqrt[3]{b}\)

\(a \ge b \Leftrightarrow \sqrt[3]{a} \ge \sqrt[3]{b}\)

\(a < b \Leftrightarrow \sqrt[3]{a} > \sqrt[3]{b}\)

\(\sqrt[3]{{ab}} = \sqrt a .\sqrt b \)

\(\dfrac{{\sqrt[3]{a}}}{{\sqrt[3]{b}}} = \dfrac{a}{b}\) với \(b \ne 0\)

\({\left( {\sqrt[3]{a}} \right)^3} =  - a\, khi\,a < 0\)

\(\dfrac{{\sqrt[3]{a}}}{{\sqrt[3]{b}}} = \sqrt[3]{{\dfrac{a}{b}}}\) với \(b \ne 0\)

\(\sqrt[3]{{ - 125}} =  - 25\)

\(\sqrt[3]{{ - 125}} =  - 5\)

\(\sqrt[3]{{ - 125}} = 25\)

\(\sqrt[3]{{ - 125}} = 5\)