Giải thích các bước giải:
$a)-\dfrac{2}{3}(x^3-6x)-\dfrac{4}{9}x\left(\dfrac{3}{8}x^2+\dfrac{1}{4}\right)$
$=-\dfrac{2}{3}x^3+4x-\dfrac{1}{6}x^3-\dfrac{1}{9}x$
$=-\dfrac{5}{6}x^3+\dfrac{35}{9}x$
$b)x^4y^6+\left(-\dfrac{1}{2}xy^2\right)^3.(8x-24y)$
$=x^4y^6-\dfrac{1}{8}x^3y^6(8x-24y)$
$=x^4y^6-x^4y^6+3x^3y^7$
$=3x^3y^7$