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**Hint:**We first take the factorisation of the given polynomial ${{x}^{3}}+343$ according to the identity ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$. We form the factorisation to find the simplified form of ${{x}^{3}}+343$ by replacing with $a=x;b=7$. We also verify the result with an arbitrary value of $x$.

**Complete step-by-step solution:**

The given polynomial ${{x}^{3}}+343$ is cubic expression. We consider ${{x}^{3}}$ as ${{\left( x \right)}^{3}}$ and 343 as ${{7}^{3}}$.

It’s a sum of two cube numbers. We factorise the given sum of the cubes according to the identity ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$.

We have ${{x}^{3}}+343$ and for the theorem we replace the values as $a=x;b=7$

We get \[{{x}^{3}}+343={{\left( x \right)}^{3}}+{{7}^{3}}=\left( x+7 \right)\left[ {{x}^{2}}-7x+49 \right]\].

We can see the term ${{x}^{3}}+343$ is a multiplication of two polynomials \[\left( x+7 \right)\] and \[\left( {{x}^{2}}-7x+49 \right)\].

These terms can’t be factored any more.

The factorisation of ${{x}^{3}}+343$ is \[\left( x+7 \right)\left( {{x}^{2}}-7x+49 \right)\].

Now we verify the result with an arbitrary value of $x=2$.

We have ${{x}^{3}}+343=\left( x+7 \right)\left( {{x}^{2}}-7x+49 \right)$.

The left-hand side of the equation gives ${{x}^{3}}+343={{2}^{3}}+343=8+343=351$.

The right-hand side of the equation gives

$\begin{align}

& \left( x+7 \right)\left( {{x}^{2}}-7x+49 \right) \\

& =\left( 2+7 \right)\left( {{2}^{2}}-7\times 2+49 \right) \\

& =9\times 39 \\

& =351 \\

\end{align}$

**Thus, verified the result of ${{x}^{3}}+343=\left( x+7 \right)\left( {{x}^{2}}-7x+49 \right)$.**

**Note:**We explain the process of getting ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$.

We need to find the simplified form of ${{\left( a+b \right)}^{3}}$. This is the cube of the sum of two numbers.

We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.

We need to multiply the term $\left( a+b \right)$ on both side of the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.

On the left side of the equation, we get ${{\left( a+b \right)}^{2}}\left( a+b \right)={{\left( a+b \right)}^{3}}$.

On the right side we have $\left( {{a}^{2}}+{{b}^{2}}+2ab \right)\left( a+b \right)$. We use multiplication and get

$\begin{align}

& \Rightarrow \left( {{a}^{2}}+{{b}^{2}}+2ab \right)\left( a+b \right) \\

& ={{a}^{2}}.a+a.{{b}^{2}}+2ab\times a+{{a}^{2}}.b+{{b}^{2}}.b+2ab.b \\

& ={{a}^{3}}+a{{b}^{2}}+2{{a}^{2}}b+{{a}^{2}}b+{{b}^{3}}+2a{{b}^{2}} \\

& ={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}} \\

\end{align}$

We also can take another form where

${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$.

This gives

$\begin{align}

& {{a}^{3}}+{{b}^{3}} \\

& ={{\left( a+b \right)}^{3}}-3ab\left( a+b \right) \\

& =\left( a+b \right)\left[ {{\left( a+b \right)}^{2}}-3ab \right] \\

& =\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right) \\

\end{align}$

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