Resolver y=(x-1)^3-(x-2)^3

$$y = ( x - 1 ) ^ { 3 } - ( x - 2 ) ^ { 3 }$$

Resolver y

$$y=3x^{2}-9x+7$$

Pasos da solución

Usar teorema binomial \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} para expandir \left(x-1\right)^{3}.

$$y=x^{3}-3x^{2}+3x-1-\left(x-2\right)^{3}$$

Usar teorema binomial \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} para expandir \left(x-2\right)^{3}.

$$y=x^{3}-3x^{2}+3x-1-\left(x^{3}-6x^{2}+12x-8\right)$$

Para calcular o oposto de x^{3}-6x^{2}+12x-8, calcula o oposto de cada termo.

$$y=x^{3}-3x^{2}+3x-1-x^{3}+6x^{2}-12x+8$$

Combina x^{3} e -x^{3} para obter 0.

$$y=-3x^{2}+3x-1+6x^{2}-12x+8$$

Combina -3x^{2} e 6x^{2} para obter 3x^{2}.

$$y=3x^{2}+3x-1-12x+8$$

Combina 3x e -12x para obter -9x.

$$y=3x^{2}-9x-1+8$$

Suma -1 e 8 para obter 7.

$$y=3x^{2}-9x+7$$

Resolver x (complex solution)

$$x=-\frac{\sqrt{12y-3}}{6}+\frac{3}{2}$$

$$x=\frac{\sqrt{12y-3}}{6}+\frac{3}{2}$$

Resolver x

$$x=-\frac{\sqrt{12y-3}}{6}+\frac{3}{2}$$

$$x=\frac{\sqrt{12y-3}}{6}+\frac{3}{2}\text{, }y\geq \frac{1}{4}$$