Question

Solve the following equation,$(a^2-9)x=a^3+27.$

Answer 1

$(a^2-9)x=a^3+27$

$(a^2-3^2)x=a^3+(3{)}^3$

$(a+3)(a-3)x=(a+3)(a^2+9-3a)$

$\left[\begin{array}{c}{x}^2-y^2=(x+y)(x-y)\\ x^3+y^3=(x+y)(x^2+y^2-xy)\end{array}\right]$

$x=\frac{(a+3)(a^2+9-3a)}{(a+3)(a-3)}$

$x=\frac{(a^2-3a+9)}{(a-3)}$.