How do you factor \[27{x^3} + 8{y^3}\]?

Hint: Here in this question, we have to find the factors of the given equation. If you see the equation it is in the form of \[{a^3} + {b^3}\]. We have a standard formula on this algebraic equation and it is given by \[{a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})\], hence by substituting the value of a and b we find the factors.

Complete step-by-step solution:
The equation is an algebraic equation or expression, where algebraic expression is a combination of variables and constant.

Now consider the given equation \[27{x^3} + 8{y^3}\], let we write in the exponential form. The number \[27{x^3}\] can be written as \[3x \times 3x \times 3x\] and the \[8{y^3}\]can be written as \[2y \times 2y \times 2y\], in the exponential form it is \[{\left( {2y} \right)^3}\]. The number \[27{x^3}\] is written as \[3x \times 3x \times 3x\] and in exponential form is \[{(3x)^3}\]. Therefore, the given equation is written as \[{\left( {3x} \right)^3} + {(2y)^3}\], the equation is in the form of \[{a^3} + {b^3}\].The \[{a^3} + {b^3}\]have a standard formula on this algebraic equation and it is given by \[{a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})\], here the value of a is \[3x\] and the value of b is \[2y\] . By substituting these values in the formula, we have
\[27{x^3} + 8{y^3} = {\left( {3x} \right)^3} + {(2y)^3} = (3x + 2y)({(3x)^2} - (3x)(2y) + {(2y)^2})\]
On simplifying we have
\[ \Rightarrow 27{x^3} + 8{y^3} = (3x + 2y)(9{x^2} - 6xy + 4{y^2})\]
The second term can’t be simplified further. Since it contains the two terms which are unknown. Therefore we keep the second term as it is.
Therefore, the factors of \[27{x^3} + 8{y^3}\] is \[(3x + 2y)(9{x^2} - 6xy + 4{y^2})\]


Note: To find the factors for algebraic equations or expressions, it depends on the degree of the equation. If the equation contains a square then we have two factors. If the equation contains a cube then we have three factors. Here this equation also contains 3 factors, the two factors are imaginary.