a, Ta có: $M$ $= \left(\frac{1\:-\:x\:^3}{1-x}+\:x\right)\left(\frac{1+x^3}{1+x}-x\right)$
$= \left(\frac{-\left(x^3-1^3\right)}{-\left(x-1\right)}+\:x\right)\left(\frac{x^3+1^3}{x+1}-x\right)$
$= \left(-\frac{\left(x-1\right)\left(x^2+x+1\right)}{-\left(x-1\right)}+\:x\right)\left(\frac{\left(x+1\right)\left(x^2-x+1\right)}{x+1}-x\right)$
$= \left(x^2+x+x+1\right)\left(x^2-x-x+1\right)$
$=\left(x^2+2x+1\right)\left(x^2-2x+1\right)$
Vậy $M$ $=\left(x^2+2x+1\right)\left(x^2-2x+1\right)$
b, Ta có: $M$ $=\left(x^2+2x+1\right)\left(x^2-2x+1\right)$
⇒ $M$ $= \left(x^2+2x\cdot \:\:1+1^2\right)\cdot \left(x^2-2x\cdot \:\:1+1^2\right)$
⇒ $M$ $= \left(x+1\right)^2\left(x-1\right)^2$
Có $M$ $>$ $0$ ⇔ $\left(x+1\right)^2\left(x-1\right)^2$ $>$ $0$ ⇔ $\left(x+1\right)^2$; $\left(x-1\right)^2$ > 0
+ Để `(x+1)^2 > 0`
⇒ $x$ $<$ $-1$ hoặc $x$ $>$ $-1$(1)
+ Để `(x-1)^2 > 0`
⇒ $x$ $>$ $1$ hoặc $x$ $<$ $1$(2)
Từ (1) và (2)
⇒ $x < -1$ hoặc $-1<x<1$ hoặc $x>1$ thì $M$ $>$ $0$
Vậy $x < -1$ hoặc $-1<x<1$ hoặc $x>1$ thì $M$ $>$ $0$
