Question

Factorize the expression $x^6-7x^3-8$.

Answer 1

Step 1: Factorizing the given expression:

The given expression is $x^6-7x^3-8$.

Using middle term splitting method the expression ca be rewritten as:

$x^6-7x^3-8$$=x^6-8x^3+x^3-8$

$$\begin{gathered} =x^3\left(x^3-8\right)+1\left(x^3-8\right) \\ =\left(x^3+1\right)\left(x^3-8\right) \end{gathered}$$

Step 2: Simplifying

Further $x^3-8$ can be rewritten as:

$x^3-8=x^3-2^3$

On comparing the above expression with identity $a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)$ we get, $a=x,b=2$.

Substituting values in identity we have,

$$\begin{gathered} x^3-8=\left(x-2\right)\left(x^2+2x+2^2\right) \\ =\left(x-2\right)\left(x^2+2x+4\right) \end{gathered}$$

So, $x^6-7x^3-8=\left(x^3+1\right)\left(x-2\right)\left(x^2+2x+4\right)$

Hence, the expression, $x^6-7x^3-8=\left(x^3+1\right)\left(x-2\right)\left(x^2+2x+4\right)$.