Factorize ${{a}^{3}}+27{{b}^{3}}+8{{c}^{3}}-18abc$.
Answer
Verified638.4k+ 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
Write any three differences between metals and nonmetals class 10 social science CBSE
Amit standing on a horizontal plane finds a bird flying class 10 maths CBSE
Two circles of radii 5 cm and 3 cm intersect at two class 10 maths CBSE
Solve the following i John and Jivanti together have class 10 maths CBSE
What is the relation between orthocenter circumcentre class 10 maths CBSE
Two plane mirrors are inclined at 70circ A ray incident class 10 physics CBSE
Trending doubts
Explain the Treaty of Vienna of 1815 class 10 social science CBSE
1 GB equals how many MB?
10 examples of evaporation in daily life with explanations
What is the full form of POSCO class 10 social science CBSE
Which is the hottest planet in the Solar system A Earth class 10 social science CBSE
Name any four life processes in living things class 10 biology CBSE