How do you factor \[{{x}^{3}}+27=0\]?

Hint: From the given question, we have been asked to factor the given equation \[{{x}^{3}}+27=0\]. We can factor the given equation by doing some transformations to the given equation. First we have to find one factor for the given equation. Then, we can get remaining factors for the given cubic polynomial.

Complete step-by-step solution:
From the question, we have been given that \[{{x}^{3}}+27=0\]
We know that we can write \[27\] as \[{{3}^{3}}\].
By doing this, we get the given equation as \[{{x}^{3}}+{{3}^{3}}=0\]
Now, on the left hand side of the equation, we get a standard polynomial which is in the form of \[{{a}^{3}}+{{b}^{3}}\].
For the standard polynomial l\[{{a}^{3}}+{{b}^{3}}\], we know that \[\left( a+b \right)\] is one of the factor for the given polynomial.
By comparing the coefficients, we get \[\left( x+3 \right)\] as a factor for the given cubic polynomial \[{{x}^{3}}+27=0\].
Now, from the question, we have been given that \[{{x}^{3}}+27=0\]
Now, as of process,
Add \[3{{x}^{2}},-3{{x}^{2}},9x,-9x\] on the left hand side of the given cubic polynomial.
By doing this, we get the above equation as \[{{x}^{3}}+3{{x}^{2}}-3{{x}^{2}}-9x+9x+27=0\]
Now, by taking the common terms out, we get the equation as \[{{x}^{2}}\left( x+3 \right)-3x\left( x+3 \right)+9\left( x+3 \right)=0\]
\[\Rightarrow \left( x+3 \right)\left( {{x}^{2}}-3x+9 \right)=0\]
Hence, the given cubic polynomial is factored.
As we have been already discussed above, by using some simple transformations, we get the given cubic polynomial factored.

Note: We should be well aware of the factorization process of polynomials. We should be well known about the process of factoring the cubic polynomial. Also, we should identify whether the given polynomial is in the form of a standard polynomial or not. Also, we should be very careful while finding the factors. We have a formulae in built given as ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$ .