Solve for: 3q^3=-8+(D)/(A)

Expression: $3{q}^{3}=-8+\frac{ D }{ A }$

Multiply both sides of the equation by $A$

$3{q}^{3}A=-8A+D$

Move the variable to the left-hand side and change its sign

$3{q}^{3}A+8A=D$

Factor out $A$ from the expression

$\left( 3{q}^{3}+8 \right)A=D$

Divide both sides of the equation by $3{q}^{3}+8$

$A=\frac{ D }{ 3{q}^{3}+8 }$