Solve y=a(x-b)^3+c

$$y = a ( x - b ) ^ { 3 } + c$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=-\frac{c-y}{\left(x-b\right)^{3}}\text{, }&x\neq b\\a\in \mathrm{C}\text{, }&y=c\text{ and }x=b\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(p-q\right)^{3}=p^{3}-3p^{2}q+3pq^{2}-q^{3} to expand \left(x-b\right)^{3}.

$$y=a\left(x^{3}-3x^{2}b+3xb^{2}-b^{3}\right)+c$$

Use the distributive property to multiply a by x^{3}-3x^{2}b+3xb^{2}-b^{3}.

$$y=ax^{3}-3ax^{2}b+3axb^{2}-ab^{3}+c$$

Swap sides so that all variable terms are on the left hand side.

$$ax^{3}-3ax^{2}b+3axb^{2}-ab^{3}+c=y$$

Subtract c from both sides.

$$ax^{3}-3ax^{2}b+3axb^{2}-ab^{3}=y-c$$

Combine all terms containing a.

$$\left(x^{3}-3x^{2}b+3xb^{2}-b^{3}\right)a=y-c$$

The equation is in standard form.

$$\left(x^{3}-3bx^{2}+3xb^{2}-b^{3}\right)a=y-c$$

Divide both sides by x^{3}-3x^{2}b+3xb^{2}-b^{3}.

$$\frac{\left(x^{3}-3bx^{2}+3xb^{2}-b^{3}\right)a}{x^{3}-3bx^{2}+3xb^{2}-b^{3}}=\frac{y-c}{x^{3}-3bx^{2}+3xb^{2}-b^{3}}$$

Dividing by x^{3}-3x^{2}b+3xb^{2}-b^{3} undoes the multiplication by x^{3}-3x^{2}b+3xb^{2}-b^{3}.

$$a=\frac{y-c}{x^{3}-3bx^{2}+3xb^{2}-b^{3}}$$

Divide y-c by x^{3}-3x^{2}b+3xb^{2}-b^{3}.

$$a=\frac{y-c}{\left(x-b\right)^{3}}$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{c-y}{\left(x-b\right)^{3}}\text{, }&x\neq b\\a\in \mathrm{R}\text{, }&y=c\text{ and }x=b\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(p-q\right)^{3}=p^{3}-3p^{2}q+3pq^{2}-q^{3} to expand \left(x-b\right)^{3}.

$$y=a\left(x^{3}-3x^{2}b+3xb^{2}-b^{3}\right)+c$$

Use the distributive property to multiply a by x^{3}-3x^{2}b+3xb^{2}-b^{3}.

$$y=ax^{3}-3ax^{2}b+3axb^{2}-ab^{3}+c$$

Swap sides so that all variable terms are on the left hand side.

$$ax^{3}-3ax^{2}b+3axb^{2}-ab^{3}+c=y$$

Subtract c from both sides.

$$ax^{3}-3ax^{2}b+3axb^{2}-ab^{3}=y-c$$

Combine all terms containing a.

$$\left(x^{3}-3x^{2}b+3xb^{2}-b^{3}\right)a=y-c$$

The equation is in standard form.

$$\left(x^{3}-3bx^{2}+3xb^{2}-b^{3}\right)a=y-c$$

Divide both sides by x^{3}-3x^{2}b+3xb^{2}-b^{3}.

$$\frac{\left(x^{3}-3bx^{2}+3xb^{2}-b^{3}\right)a}{x^{3}-3bx^{2}+3xb^{2}-b^{3}}=\frac{y-c}{x^{3}-3bx^{2}+3xb^{2}-b^{3}}$$

Dividing by x^{3}-3x^{2}b+3xb^{2}-b^{3} undoes the multiplication by x^{3}-3x^{2}b+3xb^{2}-b^{3}.

$$a=\frac{y-c}{x^{3}-3bx^{2}+3xb^{2}-b^{3}}$$

Divide y-c by x^{3}-3x^{2}b+3xb^{2}-b^{3}.

$$a=\frac{y-c}{\left(x-b\right)^{3}}$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=x+e^{\frac{4\pi i}{3}}a^{-\frac{1}{3}}\sqrt[3]{c-y}\text{; }b=x+a^{-\frac{1}{3}}\sqrt[3]{c-y}\text{; }b=x+e^{\frac{2\pi i}{3}}a^{-\frac{1}{3}}\sqrt[3]{c-y}\text{, }&a\neq 0\\b\in \mathrm{C}\text{, }&y=c\text{ and }a=0\end{matrix}\right.$$

Solve for b

$$\left\{\begin{matrix}b=\frac{\sqrt[3]{a}x+\sqrt[3]{c-y}}{\sqrt[3]{a}}\text{, }&a\neq 0\\b\in \mathrm{R}\text{, }&y=c\text{ and }a=0\end{matrix}\right.$$