Solve (lambda^3-alambda^2+24lambda-16=0

$$( \lambda ^ { 3 } - a \lambda ^ { 2 } + 24 \lambda - 16 = 0$$

Solve for a

$$a=\frac{\lambda ^{3}+24\lambda -16}{\lambda ^{2}}$$

$$\lambda \neq 0$$

Steps for Solving Linear Equation

Subtract 24\lambda from both sides. Anything subtracted from zero gives its negation.

$$\lambda ^{3}-a\lambda ^{2}-16=-24\lambda$$

Add 16 to both sides.

$$\lambda ^{3}-a\lambda ^{2}=-24\lambda +16$$

Subtract \lambda ^{3} from both sides.

$$-a\lambda ^{2}=-24\lambda +16-\lambda ^{3}$$

The equation is in standard form.

$$\left(-\lambda ^{2}\right)a=16-24\lambda -\lambda ^{3}$$

Divide both sides by -\lambda ^{2}.

$$\frac{\left(-\lambda ^{2}\right)a}{-\lambda ^{2}}=\frac{16-24\lambda -\lambda ^{3}}{-\lambda ^{2}}$$

Dividing by -\lambda ^{2} undoes the multiplication by -\lambda ^{2}.

$$a=\frac{16-24\lambda -\lambda ^{3}}{-\lambda ^{2}}$$

Divide -24\lambda +16-\lambda ^{3} by -\lambda ^{2}.

$$a=\lambda +\frac{24\lambda -16}{\lambda ^{2}}$$