$$a^3-2a^2+4a^3+3a^3 = 0$$
Answer
$$ \begin{matrix}a_1 = 0 & a_2 = \dfrac{ 1 }{ 4 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} a^3-2a^2+4a^3+3a^3 &= 0&& \text{simplify left side} \\[1 em]a^3-2a^2+7a^3 &= 0&& \\[1 em]8a^3-2a^2 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 8x^{3}-2x^{2} = 0 } $, first we need to factor our $ x^2 $.
$$ 8x^{3}-2x^{2} = x^2 \left( 8x-2 \right) $$$ x = 0 $ is a root of multiplicity $ 2 $.
The second root can be found by solving equation $ 8x-2 = 0$.
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