How do you factor \[64{x^3} + {y^3}\]?

Hint: In the given question, we have been asked to factorize the given polynomial, which is a combination of the sum of two variables. This polynomial is a cubic polynomial, i.e., a polynomial of degree three. But, if we see closely, the constant in the polynomial is also a cube. Hence, to simplify the value, we use the formula of difference of two cubes and factorize the given polynomial.

Formula used:
We are going to use the formula of sum of two cubes:
\[{a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)\]

Complete step-by-step answer:
The polynomial to be factored is \[64{x^3} + {y^3}\].
Clearly, this polynomial is the sum of two cubes, so we can apply the formula of sum of two cubes,
\[{a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)\]
So, \[64{x^3} + {y^3} = {\left( {4x} \right)^3} + {\left( y \right)^3} = \left( {4x + y} \right)\left( {16{x^2} - 4xy + {y^2}} \right)\].

Additional Information:
The formula for the sum of two numbers whole cubed is \[{\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)\]. While the formula for the difference of two numbers whole cubed is \[{\left( {a - b} \right)^3} = {a^3} - {b^3} - 3ab\left( {a - b} \right)\]. Finally, the formula for difference of two cubes is \[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)\].

Note: So, for solving questions of such type, we first write what has been given to us. Then we write down what we have to find. Then we write the formula which connects the two things. We have to see if the cubes are being added or subtracted, as the two things have totally different formulae and getting confused with using any one of them is going to give a wrong answer.