Question

Решите уравнения: 1)(Х+2)^3=х^3+8 2)(3х-1)^3=27х^3-1

Answer 1

$(x+2)^3=x^3+8$
$(x^2+4x+4)(x+2)=x^3+8$
$x^3+6x^2+12x+8-x^3-8=0$
$6x^2+12x=0$
$6x(x+2)=0$
${| {{x+2=0} \atop {6x=0}} \right. {| {{x=-2} \atop {x=0}} \right.$
Ответ: $-2;0$

$(3x-1)^3=27x^3-1$
$27 x^{3} -27 x^{2} +9x-1-27x^3+1=0$
$-27x^2+9x=0$
$-9x(3x-1)=0$
${ |{{-9x=0} \atop {3x-1=0}} {| {{x=0} \atop {3x=1}} \right. {| {{x=0} \atop {x= \frac{1}{3} }} \right.$
Ответ: $0; \frac{1}{3}$