y=ax^3+bx^2+cx+d-এর সমাধান করুন

$$y = a { x }^{ 3 } +b { x }^{ 2 } +cx+d$$

Solve for b

$$\left\{\begin{matrix}b=-\frac{ax^{3}+cx-y+d}{x^{2}}\text{, }&x\neq 0\\b\in \mathrm{R}\text{, }&y=d\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$ax^{3}+bx^{2}+cx+d=y$$

Subtract ax^{3} from both sides.

$$bx^{2}+cx+d=y-ax^{3}$$

Subtract cx from both sides.

$$bx^{2}+d=y-ax^{3}-cx$$

Subtract d from both sides.

$$bx^{2}=y-ax^{3}-cx-d$$

Reorder the terms.

$$bx^{2}=-ax^{3}-cx+y-d$$

The equation is in standard form.

$$x^{2}b=-ax^{3}-cx+y-d$$

Divide both sides by x^{2}.

$$\frac{x^{2}b}{x^{2}}=\frac{-ax^{3}-cx+y-d}{x^{2}}$$

Dividing by x^{2} undoes the multiplication by x^{2}.

$$b=\frac{-ax^{3}-cx+y-d}{x^{2}}$$