Factorize ${{a}^{3}}+27{{b}^{3}}+8{{c}^{3}}-18abc$.
Answer
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Hint: We have the formula ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz=\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)$, so we will convert the given equation in terms of ${{x}^{3}}$, ${{y}^{3}}$, ${{z}^{3}}$ and use the above formula to factorize the given equation.
Complete step by step answer:
Given that, ${{a}^{3}}+27{{b}^{3}}+8{{c}^{3}}-18abc$
Let us take the substitution $a=x$, then the above equation is modified as ${{x}^{3}}+27{{b}^{3}}+8{{c}^{3}}-18\left( x \right)bc$.
Now we have the term $27{{b}^{3}}$, we can write $27$ as ${{3}^{3}}$, then $27{{b}^{3}}={{3}^{3}}{{b}^{3}}={{\left( 3b \right)}^{3}}$.
Let us take the substitution $3b=y$, then the above equation is modified as
$\begin{align}
& {{x}^{3}}+{{\left( 3b \right)}^{3}}+8{{c}^{3}}-\dfrac{18}{3}\left( x \right)\left( 3b \right)c \\
& \Rightarrow {{x}^{3}}+{{y}^{3}}+8{{c}^{3}}-6\left( x \right)\left( y \right)c \\
\end{align}$
Now we have the term $8{{c}^{3}}$, we can write $8$ as ${{2}^{3}}$, then $8{{c}^{3}}={{2}^{3}}{{c}^{3}}={{\left( 2c \right)}^{3}}$.
Let us take the substitution $2c=z$, then the above equation is modified as
$\begin{align}
& {{x}^{3}}+{{y}^{3}}+{{\left( 2c \right)}^{3}}-\dfrac{6}{2}\left( x \right)\left( y \right)\left( 2c \right) \\
& \Rightarrow {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3\left( x \right)\left( y \right)\left( z \right) \\
\end{align}$
But we know that ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz=\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)$.
$\therefore {{a}^{3}}+27{{b}^{3}}+8{{c}^{3}}-18abc=\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)$
Where $x=a$, $y=3b$, $z=2c$. Substituting these values in the above equation, then we will get
$\begin{align}
& \Rightarrow {{a}^{3}}+27{{b}^{3}}+8{{c}^{3}}-18abc=\left( a+3b+2c \right)\left( {{\left( a \right)}^{2}}+{{\left( 3b \right)}^{2}}+{{\left( 2c \right)}^{2}}-\left( a \right)\left( 3b \right)-\left( 3b \right)\left( 2c \right)-\left( 2c \right)\left( a \right) \right) \\
& \Rightarrow {{a}^{3}}+27{{b}^{3}}+8{{c}^{3}}-18abc=\left( a+3b+2c \right)\left( {{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac \right) \\
\end{align}$
$\therefore $ Factors of ${{a}^{3}}+27{{b}^{3}}+8{{c}^{3}}-18abc$ are $a+3b+2c$, ${{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac$.
Note: We can also verify that $a+3b+2c$, ${{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac$ are the factors of ${{a}^{3}}+27{{b}^{3}}+8{{c}^{3}}-18abc$ by multiplying $a+3b+2c$ with ${{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac$.
First we will multiply $a$ with ${{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac$ after that we will multiply $3b$ with ${{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac$ and finally we will multiply $2c$ with${{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac$ and add them to verify.
Multiplying $a$ with ${{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac$, we will get
$\begin{align}
& a\left( {{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac \right)={{a}^{3}}+9a{{b}^{2}}+4a{{c}^{2}}-3{{a}^{2}}b-6abc-2{{a}^{2}}c \\
& \Rightarrow a\left( {{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac \right)={{a}^{3}}-3{{a}^{2}}b-2{{a}^{2}}c+9a{{b}^{2}}+4a{{c}^{2}}-6abc \\
\end{align}$
Multiplying $3b$ with ${{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac$, we will get
$\begin{align}
& 3b\left( {{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac \right)=3{{a}^{2}}b+27{{b}^{3}}+12b{{c}^{2}}-9a{{b}^{2}}-18{{b}^{2}}c-6abc \\
& \Rightarrow 3b\left( {{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac \right)=27{{b}^{3}}+3{{a}^{2}}b-9a{{b}^{2}}-18{{b}^{2}}c+12b{{c}^{2}}-6abc \\
\end{align}$
Multiplying $2c$ with ${{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac$, we will get
$\begin{align}
& 2c\left( {{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac \right)=2{{a}^{2}}c+18c{{b}^{2}}+8{{c}^{3}}-6abc-12b{{c}^{2}}-4a{{c}^{2}} \\
& \Rightarrow 2c\left( {{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac \right)=8{{c}^{3}}+2{{a}^{2}}c+18c{{b}^{2}}-12b{{c}^{2}}-4a{{c}^{2}}-6abc \\
\end{align}$
Now adding all the above terms, then we will get
$\begin{align}
& a\left( {{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac \right)+3b\left( {{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac \right)+2c\left( {{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac \right) \\
& =\left( {{a}^{3}}-3{{a}^{2}}b-2{{a}^{2}}c+9a{{b}^{2}}+4a{{c}^{2}}-6abc \right)+\left( 27{{b}^{3}}+3{{a}^{2}}b-9a{{b}^{2}}-18{{b}^{2}}c+12b{{c}^{2}}-6abc \right)+\left( 8{{c}^{3}}+2{{a}^{2}}c+18c{{b}^{2}}-12b{{c}^{2}}-4a{{c}^{2}}-6abc \right) \\
& \Rightarrow \left( a+3b+2c \right)\left( {{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac \right) \\
& =\left( {{a}^{3}}+27{{b}^{3}}+8{{c}^{3}} \right)+\left( -3{{a}^{2}}b+3{{a}^{2}}b \right)+\left( -2{{a}^{2}}c+2{{a}^{2}}c \right)+\left( 9a{{b}^{2}}-9a{{b}^{2}} \right)+\left( 4a{{c}^{2}}-4a{{c}^{2}} \right)+\left( -18{{b}^{2}}c+18c{{b}^{2}} \right)+\left( 2b{{c}^{2}}-12b{{c}^{2}} \right)+\left( -6abc-6abc-6abc \right) \\
& \Rightarrow \left( a+3b+2c \right)\left( {{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac \right)={{a}^{3}}+27{{b}^{3}}+8{{c}^{3}}-18abc \\
\end{align}$
Hence Verified.
Complete step by step answer:
Given that, ${{a}^{3}}+27{{b}^{3}}+8{{c}^{3}}-18abc$
Let us take the substitution $a=x$, then the above equation is modified as ${{x}^{3}}+27{{b}^{3}}+8{{c}^{3}}-18\left( x \right)bc$.
Now we have the term $27{{b}^{3}}$, we can write $27$ as ${{3}^{3}}$, then $27{{b}^{3}}={{3}^{3}}{{b}^{3}}={{\left( 3b \right)}^{3}}$.
Let us take the substitution $3b=y$, then the above equation is modified as
$\begin{align}
& {{x}^{3}}+{{\left( 3b \right)}^{3}}+8{{c}^{3}}-\dfrac{18}{3}\left( x \right)\left( 3b \right)c \\
& \Rightarrow {{x}^{3}}+{{y}^{3}}+8{{c}^{3}}-6\left( x \right)\left( y \right)c \\
\end{align}$
Now we have the term $8{{c}^{3}}$, we can write $8$ as ${{2}^{3}}$, then $8{{c}^{3}}={{2}^{3}}{{c}^{3}}={{\left( 2c \right)}^{3}}$.
Let us take the substitution $2c=z$, then the above equation is modified as
$\begin{align}
& {{x}^{3}}+{{y}^{3}}+{{\left( 2c \right)}^{3}}-\dfrac{6}{2}\left( x \right)\left( y \right)\left( 2c \right) \\
& \Rightarrow {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3\left( x \right)\left( y \right)\left( z \right) \\
\end{align}$
But we know that ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz=\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)$.
$\therefore {{a}^{3}}+27{{b}^{3}}+8{{c}^{3}}-18abc=\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)$
Where $x=a$, $y=3b$, $z=2c$. Substituting these values in the above equation, then we will get
$\begin{align}
& \Rightarrow {{a}^{3}}+27{{b}^{3}}+8{{c}^{3}}-18abc=\left( a+3b+2c \right)\left( {{\left( a \right)}^{2}}+{{\left( 3b \right)}^{2}}+{{\left( 2c \right)}^{2}}-\left( a \right)\left( 3b \right)-\left( 3b \right)\left( 2c \right)-\left( 2c \right)\left( a \right) \right) \\
& \Rightarrow {{a}^{3}}+27{{b}^{3}}+8{{c}^{3}}-18abc=\left( a+3b+2c \right)\left( {{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac \right) \\
\end{align}$
$\therefore $ Factors of ${{a}^{3}}+27{{b}^{3}}+8{{c}^{3}}-18abc$ are $a+3b+2c$, ${{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac$.
Note: We can also verify that $a+3b+2c$, ${{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac$ are the factors of ${{a}^{3}}+27{{b}^{3}}+8{{c}^{3}}-18abc$ by multiplying $a+3b+2c$ with ${{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac$.
First we will multiply $a$ with ${{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac$ after that we will multiply $3b$ with ${{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac$ and finally we will multiply $2c$ with${{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac$ and add them to verify.
Multiplying $a$ with ${{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac$, we will get
$\begin{align}
& a\left( {{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac \right)={{a}^{3}}+9a{{b}^{2}}+4a{{c}^{2}}-3{{a}^{2}}b-6abc-2{{a}^{2}}c \\
& \Rightarrow a\left( {{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac \right)={{a}^{3}}-3{{a}^{2}}b-2{{a}^{2}}c+9a{{b}^{2}}+4a{{c}^{2}}-6abc \\
\end{align}$
Multiplying $3b$ with ${{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac$, we will get
$\begin{align}
& 3b\left( {{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac \right)=3{{a}^{2}}b+27{{b}^{3}}+12b{{c}^{2}}-9a{{b}^{2}}-18{{b}^{2}}c-6abc \\
& \Rightarrow 3b\left( {{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac \right)=27{{b}^{3}}+3{{a}^{2}}b-9a{{b}^{2}}-18{{b}^{2}}c+12b{{c}^{2}}-6abc \\
\end{align}$
Multiplying $2c$ with ${{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac$, we will get
$\begin{align}
& 2c\left( {{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac \right)=2{{a}^{2}}c+18c{{b}^{2}}+8{{c}^{3}}-6abc-12b{{c}^{2}}-4a{{c}^{2}} \\
& \Rightarrow 2c\left( {{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac \right)=8{{c}^{3}}+2{{a}^{2}}c+18c{{b}^{2}}-12b{{c}^{2}}-4a{{c}^{2}}-6abc \\
\end{align}$
Now adding all the above terms, then we will get
$\begin{align}
& a\left( {{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac \right)+3b\left( {{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac \right)+2c\left( {{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac \right) \\
& =\left( {{a}^{3}}-3{{a}^{2}}b-2{{a}^{2}}c+9a{{b}^{2}}+4a{{c}^{2}}-6abc \right)+\left( 27{{b}^{3}}+3{{a}^{2}}b-9a{{b}^{2}}-18{{b}^{2}}c+12b{{c}^{2}}-6abc \right)+\left( 8{{c}^{3}}+2{{a}^{2}}c+18c{{b}^{2}}-12b{{c}^{2}}-4a{{c}^{2}}-6abc \right) \\
& \Rightarrow \left( a+3b+2c \right)\left( {{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac \right) \\
& =\left( {{a}^{3}}+27{{b}^{3}}+8{{c}^{3}} \right)+\left( -3{{a}^{2}}b+3{{a}^{2}}b \right)+\left( -2{{a}^{2}}c+2{{a}^{2}}c \right)+\left( 9a{{b}^{2}}-9a{{b}^{2}} \right)+\left( 4a{{c}^{2}}-4a{{c}^{2}} \right)+\left( -18{{b}^{2}}c+18c{{b}^{2}} \right)+\left( 2b{{c}^{2}}-12b{{c}^{2}} \right)+\left( -6abc-6abc-6abc \right) \\
& \Rightarrow \left( a+3b+2c \right)\left( {{a}^{2}}+9{{b}^{2}}+4{{c}^{2}}-3ab-6bc-2ac \right)={{a}^{3}}+27{{b}^{3}}+8{{c}^{3}}-18abc \\
\end{align}$
Hence Verified.
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