Question

$3\sqrt{3}{{x}^{3}}+{{y}^{3}}$is equal to
A. \[\left( \sqrt{3}x+y \right)\left( 3{{x}^{2}}-\sqrt{3}xy+{{y}^{2}} \right)\]
B. \[\left( \sqrt{3}x-y \right)\left( 3{{x}^{2}}+\sqrt{3}xy+{{y}^{2}} \right)\]
C. \[\left( \sqrt{3}x+y \right)\left( 3{{x}^{2}}+\sqrt{3}xy+{{y}^{2}} \right)\]
D. \[\left( \sqrt{3}x-y \right)\left( 3{{x}^{2}}-\sqrt{3}xy-{{y}^{2}} \right)\]

Answer 1

Hint: First change the given algebraic expression into the form \[{{a}^{3}}+{{b}^{3}}\] and then use the identity $''{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)''$ to break it into its factors to get the answer.

Complete step by step answer:
Given algebraic expression $3\sqrt{3}{{x}^{3}}+{{y}^{3}}$.
We have to find an expression among the options which is equivalent to this given expression.
Among the options, we can see that all the expressions in the options are simpler expressions multiplied together.
Its means we have to convert $3\sqrt{3}{{x}^{3}}+{{y}^{3}}$ into product of simpler expression i.e. we have to factorise this expression.
We know the identity: ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$
To use this identity, first we have to convert the given expression into the form \[{{a}^{3}}+{{b}^{3}}\].
$3\sqrt{3}{{x}^{3}}$ can be written as \[{{\left( \sqrt{3}x \right)}^{3}}\]as,
$\sqrt{3}x\times \sqrt{3}x\times \sqrt{3}x=3\sqrt{3}{{x}^{3}}$ and
${{y}^{3}}$ can be written as ${{\left( y \right)}^{3}}$as,
$y\times y\times y={{y}^{3}}$
So, $3\sqrt{3}{{x}^{3}}+{{y}^{3}}={{\left( \sqrt{3}x \right)}^{3}}+{{\left( y \right)}^{3}}$
Putting $a=\sqrt{3}x\ and\ b=y$ in the above identity $''{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)''$, we will get,
\[\begin{align}
  & {{\left( \sqrt{3}x \right)}^{3}}+{{\left( y \right)}^{3}}=\left[ \left( \sqrt{3}x \right)+\left( y \right) \right]\left[ {{\left( \sqrt{3}x \right)}^{2}}-\left( \sqrt{3}x \right)\left( y \right)+{{\left( y \right)}^{2}} \right] \\
 & \Rightarrow 3\sqrt{3}{{x}^{3}}+{{y}^{3}}=\left( \sqrt{3}x+y \right)\left( 3{{x}^{2}}-\sqrt{3}xy+{{y}^{2}} \right) \\
\end{align}\]
Hence, $\left( 3\sqrt{3}{{x}^{3}}+{{y}^{3}} \right)$ will be equivalent to \[\left( \sqrt{3}x+y \right)\left( 3{{x}^{2}}-\sqrt{3}xy+{{y}^{2}} \right)\]

So, the correct answer is “Option A”.

Note: We can prove the identity used in the following way:
Identity: ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$
Proof: $LHS={{a}^{3}}+{{b}^{3}}$
Adding and subtracting $3ab\left( a+b \right)$, we will get,
$LHS={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)-3ab\left( a+b \right)$
We know ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$
So, replacing ${{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)\ by\ {{\left( a+b \right)}^{3}}$, we will get,
\[LHS={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)\]
Taking \[\left( a+b \right)\] common from both the terms, we will get,
\[\begin{align}
  & LHS=\left( a+b \right)\left[ {{\left( a+b \right)}^{2}}-3ab \right] \\
 & We\ know\ {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\
 & \Rightarrow LHS=\left( a+b \right)\left[ {{a}^{2}}+{{b}^{2}}+2ab-3ab \right] \\
 & \Rightarrow LHS=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right) \\
 & \Rightarrow LHS=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right) \\
 & \Rightarrow LHS=RHS \\
\end{align}\]
Hence, Proved.