Question

How do you factor ${x^8} - {y^8}$?

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 of difference of two squares and simplify the given expression using it. First of all convert the given expression in the form of the difference of two squares.

Complete step-by-step solution:
Take the given expression: ${x^8} - {y^8}$
It can be re-written as –
$\Rightarrow {({x^4})^2} - {({y^4})^2}$
Now, we can use here the difference of two squares identity like $(a - b)(a + b) = {a^2} - {b^2}$
$\Rightarrow ({x^4} + {y^4})({x^4} - {y^4})$
Again, the above expression can be re-written as the difference of two squares –
$\Rightarrow ({x^4} + {y^4})[{({x^2})^2} - {({y^2})^2}]$
Place the identity for the difference of two squares –
$\Rightarrow ({x^4} + {y^4})({x^2} + {y^2})({x^2} - {y^2})$
Again, the term multiple can be re-written as difference of squares –
$\Rightarrow ({x^4} + {y^4})({x^2} + {y^2})[{(x)^2} - ({y^2})]$
Apply formula in the $(a - b)(a + b) = {a^2} - {b^2}$ formula we get the expression as –
$\Rightarrow ({x^4} + {y^4})({x^2} + {y^2})(x + y)(x - y)$
This is the required solution.

Thus the required solution is $({x^4} + {y^4})({x^2} + {y^2})(x + y)(x - y)$.

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