Factorize ${{a}^{3}}+27{{b}^{3}}+8{{c}^{3}}-18abc$.
Answer
Verified634.5k+ views
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.
Recently Updated Pages
Master Class 10 Computer Science: Engaging Questions & Answers for Success
Differentiate between voluntary action and reflex class 10 biology CBSE
The uses of bleaching powder are A It is used bleaching class 10 chemistry CBSE
Fill in the blanks with abstract nouns of the words class 10 english CBSE
How many threedigit numbers are there class 10 maths CBSE
What is a reflex arc class 10 biology CBSE
Trending doubts
Explain the Treaty of Vienna of 1815 class 10 social science CBSE
10 examples of evaporation in daily life with explanations
What is the full form of POSCO class 10 social science CBSE
Make a sketch of the human nerve cell What function class 10 biology CBSE
Identify the feminine form of noun nephew a shenephew class 10 english CBSE
Compare the advantages and disadvantages of multipurpose class 10 social science CBSE