Question

How do you factor $ {x^6} - 2{x^3} + 1 $ ?

Answer 1

Hint: As we know that factorising is the reverse of expanding brackets, it is an important way of solving equations. The first step of factorising an expression is to take out any common factors which the terms have. So if we were asked to factor the expression $ {x^2} + x $ , since $ x $ goes into both terms, we would write $ x(x + 1) $ . Here we will use identities which will help us to factorise an algebraic expression easily i.e. $ {(a - b)^2} = {a^2} - 2ab + {b^2} $ and $ {a^2} - {b^2} = (a + b)(a - b) $

Complete step-by-step answer:
We can rewrite $ {x^6} $ as $ {({x^3})^2} $ . So the equation will be
$\Rightarrow {({x^3})^2} - 2{x^3} + 1 $ .
Let $ u = {x^3} $ . Now substitute $ u $ for all occurrences of $ {x^3} $ i.e. $ {u^2} - 2u + 1 $ Here we apply the perfect square formula which gives us $ {(u - 1)^2} $ . Again replace all occurrences of $ u $ with $ {x^3} $ .
$ {({x^3} - 1)^2} $ We can write $ 1 $ as $
{1^3} $ $ \Rightarrow {({x^3} - {1^3})^2} $ .
Since both terms are perfect cubes, we can factor using the difference of cubes formula, $ {a^3} - {b^3} = (a - b)({a^2} + ab + {b^2}) $ where
$ a = x $ and $ b = 1 $ . So it is,
$ {x^3} - 1 = $ $ \{ (x - 1)({x^2} + x\times 1 + {1^2})\} $
$
{x^6} - 2{x^3} + 1 = {\{ (x - 1)({x^2} + x + 1)\} ^2} \\
\Rightarrow (x - 1)({x^2} + x + 1)(x - 1)({x^2} + x + 1)
$
 further it can also be written as
$ {(x - 1)^2}{({x^2} + x + 1)^2} $ .
Hence the answer is $ {(x - 1)^2}{({x^2} + x + 1)^2} $ .
So, the correct answer is “$ {(x - 1)^2}{({x^2} + x + 1)^2} $”.

Note: We should keep in mind while solving these expressions that we use correct identities to factorise the given algebraic expressions and keep checking the negative and positive sign otherwise it will give the wrong answer. $ {x^2} + x + 1 $ has no linear factors with real coefficients. To check this we can apply the discriminant formula and being negative discriminant it has no real roots. These are some of the standard algebraic identities. This is as far we can go with real coefficients as the remaining quadratic factors all have complex zeros.