Question

How do you factor ${x^3} - 16?$

Answer 1

Hint: Factorization of any polynomials can be written as the product of its factors having the degree less than or equal to the original polynomial. Here we will use the identity for the difference of two squares, difference of two cubes and then will simplify for the resultant required solution.

Complete step-by-step solution:
Take the given expression: ${x^3} - 16$
Above expression can be re-written as: ${(x)^3} - {(2\sqrt[3]{2})^3}$
Now, using the difference of cubes identity which can be given as: ${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$
$\Rightarrow [(x - 2)[{(x)^2} + (x)(2\sqrt[3]{2}) + {(2\sqrt[3]{2})^2}]$
Simplify the above equation.
$\Rightarrow [(x - 2)({x^2} + 2\sqrt[3]{2}x + 4{(2)^{\dfrac{2}{3}}}]$

Hence, the required solution –${x^3} - 16 = [(x - 2)({x^2} + 2\sqrt[3]{2}x + 4{(2)^{\dfrac{2}{3}}}]$

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 ${2^2} = 2 \times 2 = 4$. 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 ${2^3} = 2 \times 2 \times 2 = 8$.