Solve Deltay=(4+Deltat)^3+3-(4^3+3)

$$\Delta y = ( 4 + \Delta t ) ^ { 3 } + 3 - ( 4 ^ { 3 } + 3 )$$

Solve for y

$$\left\{\begin{matrix}y=t\left(12t\Delta +\left(t\Delta \right)^{2}+48\right)\text{, }&\Delta \neq 0\\y\in \mathrm{R}\text{, }&\Delta =0\end{matrix}\right.$$

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(4+\Delta t\right)^{3}.

$$\Delta y=64+48\Delta t+12\Delta ^{2}t^{2}+\Delta ^{3}t^{3}+3-\left(4^{3}+3\right)$$

Add 64 and 3 to get 67.

$$\Delta y=67+48\Delta t+12\Delta ^{2}t^{2}+\Delta ^{3}t^{3}-\left(4^{3}+3\right)$$

Calculate 4 to the power of 3 and get 64.

$$\Delta y=67+48\Delta t+12\Delta ^{2}t^{2}+\Delta ^{3}t^{3}-\left(64+3\right)$$

Add 64 and 3 to get 67.

$$\Delta y=67+48\Delta t+12\Delta ^{2}t^{2}+\Delta ^{3}t^{3}-67$$

Subtract 67 from 67 to get 0.

$$\Delta y=48\Delta t+12\Delta ^{2}t^{2}+\Delta ^{3}t^{3}$$

The equation is in standard form.

$$\Delta y=t^{3}\Delta ^{3}+12t^{2}\Delta ^{2}+48t\Delta$$

Divide both sides by \Delta .

$$\frac{\Delta y}{\Delta }=\frac{t\Delta \left(t^{2}\Delta ^{2}+12t\Delta +48\right)}{\Delta }$$

Dividing by \Delta undoes the multiplication by \Delta .

$$y=\frac{t\Delta \left(t^{2}\Delta ^{2}+12t\Delta +48\right)}{\Delta }$$

Divide \Delta t\left(48+12\Delta t+\Delta ^{2}t^{2}\right) by \Delta .

$$y=t\left(t^{2}\Delta ^{2}+12t\Delta +48\right)$$

Solve for t

$$\left\{\begin{matrix}t=\frac{\sqrt[3]{y\Delta +64}-4}{\Delta }\text{, }&\Delta \neq 0\\t\in \mathrm{R}\text{, }&\Delta =0\end{matrix}\right.$$