Question

a
${a}^{6} - 7 {a}^{3} - 8$

Answer 1

We have to factor $a^6-7a^3-8$

there is no direct method for higher degree polynomials so we may need to rearranges the given problem so that it gets converted into quadratic or cubic polynomial for which we know how to factor.

Let's rewrite that as:

$=(a^3)^2-7(a^3)-8$

Replace $a^3$ by some other variable say x. then problem changes to

$=x^2-7x-8$

Now we can easily factor this quadratic polynomial

$=x^2-8x+1x-8$

$=x(x-8)+1(x-8)$

$=(x+1)(x-8)$

plug value of $x=a^3$

$=(a^3+1)(a^3-8)$

$=(a^3+1^3)(a^3-2^3)$

Factor using formula $x^3+y^3=(x+y)(x^2-xy+y^2)$

$=(a+1)(a^2-a*1+1^2)(a^3-2^3)$

$=(a+1)(a^2-a+1)(a^3-2^3)$

Factor using formula $x^3-y^3=(x-y)(x^2+xy+y^2)$

$=(a+1)(a^2-a+1)(a-2)(a^2+a*2+2^2)$

$=(a+1)(a^2-a+1)(a-2)(a^2+2a+4)$

Hence final answer is $(a+1)(a^2-a+1)(a-2)(a^2+2a+4)$