How do you find the quotient of $\dfrac{{{x^3} + 27}}{{x + 3}}$?

Hint: Here, we will use the identity of the sum of cubes of two terms and then find the common factor from the numerator and the denominator and then the resultant term is quotient.

Complete step by step solution:
Take the given expression: $\dfrac{{{x^3} + 27}}{{x + 3}}$
Use the identity of the sum of the cubes as ${a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})$in the above equation.
Here, $a = x$ and $b = 3$
 \[{(x)^3} + {(3)^3} = (x + 3)\left( {{{(x)}^2} - (x)(3) + {{(3)}^2}} \right)\]
Simplify the above expression –
 \[{(x)^3} + {(3)^3} = (x + 3)\left( {{x^2} - 3x + 9} \right)\]
Place the above value in the given expression –
 \[\dfrac{{{x^3} + 27}}{{x + 3}} = \dfrac{{(x + 3)\left( {{x^2} - 3x + 9} \right)}}{{(x + 3)}}\]
Common factors from the numerator and the denominator cancels each other. Therefore, remove $(x + 3)$from the numerator and the denominator.
Using the Division Algorithm which states that –
Dividend $ = $Divisor $ \times $Quotient $ + $Remainder
So, it applies that Quotient is \[\left( {{x^2} - 3x + 9} \right)\]
So, the correct answer is “ \[\left( {{x^2} - 3x + 9} \right)\] ”.

Note: Know the concepts of squares and cubes. Square is the number multiplied itself and cube it the number multiplied thrice. Square is the product of same number twice such as ${n^2} = n \times n$ for Example square of $2$ is ${2^2} = 2 \times 2$ simplified form of squared number is ${2^2} = 2 \times 2 = 4$ and square-root is denoted by $\sqrt {{n^2}} = \sqrt {n \times n} $ For Example: $\sqrt {{2^2}} = \sqrt 4 = 2$ Similarly cube is the product of same number three times such as ${n^3} = n \times n \times n$ for Example cube of $2$ is ${2^3} = 2 \times 2 \times 2$ simplified form of cubed number is ${2^3} = 2 \times 2 \times 2 = 8$. and cube-root is denoted by $\sqrt[3] {{{n^3}}} = \sqrt {n \times n \times n} = n$ For Example: $\sqrt[3] {8} = \sqrt[3] {{{2^3}}} = 2$ Do not be confused in square and square-root similarly cubes and cube-root, know the concepts properly and apply accordingly.