Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
DKXD:\,\,\,a \ne - b\\
a,\\
S = \dfrac{{{a^3} + 2a{b^2} + 3{b^3}}}{{{{\left( {a + b} \right)}^2}}}\\
= \dfrac{{\left( {{a^3} + {a^2}b} \right) + \left( { - {a^2}b - a{b^2}} \right) + \left( {3a{b^2} + 3{b^3}} \right)}}{{{{\left( {a + b} \right)}^2}}}\\
= \dfrac{{{a^2}.\left( {a + b} \right) - ab.\left( {a + b} \right) + 3{b^2}\left( {a + b} \right)}}{{{{\left( {a + b} \right)}^2}}}\\
= \dfrac{{\left( {a + b} \right)\left( {{a^2} - ab + 3{b^2}} \right)}}{{{{\left( {a + b} \right)}^2}}}\\
= \dfrac{{{a^2} - ab + 3{b^2}}}{{a + b}}\\
b,\\
S = 3b\\
\Leftrightarrow \dfrac{{{a^2} - ab + 3{b^2}}}{{a + b}} = 3b\\
\Leftrightarrow {a^2} - ab + 3{b^2} = 3b.\left( {a + b} \right)\\
\Leftrightarrow {a^2} - ab + 3{b^2} = 3ab + 3{b^2}\\
\Leftrightarrow {a^2} - 4ab = 0\\
\Leftrightarrow a.\left( {a - 4b} \right) = 0\\
a \ne 0 \Rightarrow a - 4b = 0\\
\Leftrightarrow a = 4b
\end{array}\)
