How do you factor \[{{x}^{3}}-729\]?

Hint: To solve the given question we will need the given properties and formula. If \[{{a}^{3}}={{b}^{3}}\] then taking the cube root of both sides we get, \[a=b\]. We should also remember the following factorization formula which states that, \[{{x}^{3}}-{{y}^{3}}=(x-y)({{x}^{2}}+xy+{{y}^{2}})\]. This is called the difference of cubes formula.

Complete step by step answer:
The given expression is \[{{x}^{3}}-729\], substituting \[{{y}^{3}}\] at place of 729 in the expression we get, \[{{x}^{3}}-{{y}^{3}}\].
Because of the substitution, we can say that \[{{y}^{3}}=729\] .
We know that 729 is the cube of 9, so 729 can also be written as \[{{9}^{3}}\], the above expression becomes,
\[\Rightarrow {{y}^{3}}={{9}^{3}}\]
Taking the cube root of both sides of the above expression we get,
\[\Rightarrow {{\left( {{y}^{3}} \right)}^{\dfrac{1}{3}}}={{\left( 729 \right)}^{\dfrac{1}{3}}}\]
\[\Rightarrow y=9\]
Hence, we get the value of y as 9, substituting this value in the expression \[{{x}^{3}}-{{y}^{3}}\] we get \[{{x}^{3}}-{{9}^{3}}\]
As both terms in this expression are perfect cubes, we can use the difference of cubes formula which states that \[{{x}^{3}}-{{y}^{3}}=(x-y)({{x}^{2}}+xy+{{y}^{2}})\].
Using this formula, the expression \[{{x}^{3}}-{{9}^{3}}\] can be factorized as follows. Comparing with the standard formula \[{{x}^{3}}-{{y}^{3}}\] here y is replaced with 9 so the factorization will be,
\[\Rightarrow (x-9)({{x}^{2}}+x\times 9+{{9}^{2}})\]
The above expression can also be written as \[(x-9)({{x}^{2}}+9x+81)\].

Hence, the factorization of \[{{x}^{3}}-729\] is \[(x-9)({{x}^{2}}+9x+81)\].

Note: These types of problems are not very difficult and can be easily solved by remembering formulas like difference of cubes: \[{{x}^{3}}-{{y}^{3}}=(x-y)({{x}^{2}}+xy+{{y}^{2}})\], sum of cubes: \[{{x}^{3}}+{{y}^{3}}=(x+y)({{x}^{2}}-xy+{{y}^{2}})\], difference of squares: \[{{x}^{2}}-{{y}^{2}}=(x+y)(x-y)\], etc.
It should be noted that these formulas can be used even if numbers are not perfect cubes or perfect squares, but in that case, we will not get integral values.