Solve left(z+mright)^3+left(z-mright)^3=2left(z^3+6m^3right)

$${ \left(z+m \right) }^{ 3 } + { \left(z-m \right) }^{ 3 } = 2 \left( { z }^{ 3 } +6 { m }^{ 3 } \right)$$

Solve for z (complex solution)

$$\left\{\begin{matrix}\\z=2m\text{, }&\text{unconditionally}\\z\in \mathrm{C}\text{, }&m=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(z+m\right)^{3}.

$$z^{3}+3z^{2}m+3zm^{2}+m^{3}+\left(z-m\right)^{3}=2\left(z^{3}+6m^{3}\right)$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(z-m\right)^{3}.

$$z^{3}+3z^{2}m+3zm^{2}+m^{3}+z^{3}-3z^{2}m+3zm^{2}-m^{3}=2\left(z^{3}+6m^{3}\right)$$

Combine z^{3} and z^{3} to get 2z^{3}.

$$2z^{3}+3z^{2}m+3zm^{2}+m^{3}-3z^{2}m+3zm^{2}-m^{3}=2\left(z^{3}+6m^{3}\right)$$

Combine 3z^{2}m and -3z^{2}m to get 0.

$$2z^{3}+3zm^{2}+m^{3}+3zm^{2}-m^{3}=2\left(z^{3}+6m^{3}\right)$$

Combine 3zm^{2} and 3zm^{2} to get 6zm^{2}.

$$2z^{3}+6zm^{2}+m^{3}-m^{3}=2\left(z^{3}+6m^{3}\right)$$

Combine m^{3} and -m^{3} to get 0.

$$2z^{3}+6zm^{2}=2\left(z^{3}+6m^{3}\right)$$

Use the distributive property to multiply 2 by z^{3}+6m^{3}.

$$2z^{3}+6zm^{2}=2z^{3}+12m^{3}$$

Subtract 2z^{3} from both sides.

$$2z^{3}+6zm^{2}-2z^{3}=12m^{3}$$

Combine 2z^{3} and -2z^{3} to get 0.

$$6zm^{2}=12m^{3}$$

The equation is in standard form.

$$6m^{2}z=12m^{3}$$

Divide both sides by 6m^{2}.

$$\frac{6m^{2}z}{6m^{2}}=\frac{12m^{3}}{6m^{2}}$$

Dividing by 6m^{2} undoes the multiplication by 6m^{2}.

$$z=\frac{12m^{3}}{6m^{2}}$$

Divide 12m^{3} by 6m^{2}.

$$z=2m$$

Solve for z

$$\left\{\begin{matrix}\\z=2m\text{, }&\text{unconditionally}\\z\in \mathrm{R}\text{, }&m=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(z+m\right)^{3}.

$$z^{3}+3z^{2}m+3zm^{2}+m^{3}+\left(z-m\right)^{3}=2\left(z^{3}+6m^{3}\right)$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(z-m\right)^{3}.

$$z^{3}+3z^{2}m+3zm^{2}+m^{3}+z^{3}-3z^{2}m+3zm^{2}-m^{3}=2\left(z^{3}+6m^{3}\right)$$

Combine z^{3} and z^{3} to get 2z^{3}.

$$2z^{3}+3z^{2}m+3zm^{2}+m^{3}-3z^{2}m+3zm^{2}-m^{3}=2\left(z^{3}+6m^{3}\right)$$

Combine 3z^{2}m and -3z^{2}m to get 0.

$$2z^{3}+3zm^{2}+m^{3}+3zm^{2}-m^{3}=2\left(z^{3}+6m^{3}\right)$$

Combine 3zm^{2} and 3zm^{2} to get 6zm^{2}.

$$2z^{3}+6zm^{2}+m^{3}-m^{3}=2\left(z^{3}+6m^{3}\right)$$

Combine m^{3} and -m^{3} to get 0.

$$2z^{3}+6zm^{2}=2\left(z^{3}+6m^{3}\right)$$

Use the distributive property to multiply 2 by z^{3}+6m^{3}.

$$2z^{3}+6zm^{2}=2z^{3}+12m^{3}$$

Subtract 2z^{3} from both sides.

$$2z^{3}+6zm^{2}-2z^{3}=12m^{3}$$

Combine 2z^{3} and -2z^{3} to get 0.

$$6zm^{2}=12m^{3}$$

The equation is in standard form.

$$6m^{2}z=12m^{3}$$

Divide both sides by 6m^{2}.

$$\frac{6m^{2}z}{6m^{2}}=\frac{12m^{3}}{6m^{2}}$$

Dividing by 6m^{2} undoes the multiplication by 6m^{2}.

$$z=\frac{12m^{3}}{6m^{2}}$$

Divide 12m^{3} by 6m^{2}.

$$z=2m$$

Solve for m

$$m=0$$

$$m=\frac{z}{2}$$