a: Đặt x+2=a; x+3=b
Theo đề, ta có pt: \(\left(a+b\right)^3-a^3-b^3=0\)
=>-3ab(a+b)=0
=>ab(a+b)=0
=>(2x+5)(x+2)(x+3)=0
hay \(x\in\left\{-\dfrac{5}{2};-3;-2\right\}\)
b: Đặt x+7=a; x+6=b
Theo đề, ta có pt: \(a^3+b^3-\left(a+b\right)^3=0\)
=>ab(a+b)=0
=>(x+7)(2x+13)(x+6)=0
hay \(x\in\left\{-7;-6;-\dfrac{13}{2}\right\}\)