Solve the equation a^3-4a^2+4a=0 The solutions are a=square and a= square

To solve the equation $a^{3}-4a^{2}+4a=0$, we can first factor out the common factor of $a$ from each term:<br /><br />$a(a^{2}-4a+4)=0$<br /><br />Now, we have two factors: $a$ and $(a^{2}-4a+4)$. We can set each factor equal to zero and solve for $a$:<br /><br />$a=0$<br /><br />$(a^{2}-4a+4)=0$<br /><br />For the quadratic equation $(a^{2}-4a+4)=0$, we can use the quadratic formula to find the solutions:<br /><br />$a=\frac{-(-4)\pm\sqrt{(-4)^{2}-4(1)(4)}}{2(1)}$<br /><br />$a=\frac{4\pm\sqrt{16-16}}{2}$<br /><br />$a=\frac{4\pm0}{2}$<br /><br />$a=2$<br /><br />So, the solutions to the equation $a^{3}-4a^{2}+4a=0$ are $a=0$ and $a=2$.