Question

9-a^6+2a^3b^3-b^6 factorize

Answer 1

Answer:

$[3-(a^3-b^3)][3+(a^3-b^3)]$

Step-by-step explanation:

$9-a^6+2a^3b^3-b^6\\=9-(a^6-2a^3b^3+b^6)\\=3^2-((a^3)^2-2a^3b^3+(b^3)^2)$.

Notice that the bracket $((a^3)^2-2a^3b^3+(b^3)^2)$ is similar to ($a^2-2ab+b^2$), which is factorized as: $(a-b)^2$.

Therefore, if we follow the same procedure:

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

Notice that

$3^2-(a^3-b^3)^2$ is similar to the difference between two squares formula: $a^2-b^2=(a-b)(a+b)$

Therefore,

$3^2-(a^3-b^3)^2\\=[3-(a^3-b^3)][3+(a^3-b^3)]$

Answer 2

Solution:

To factorise

$9 - {a}^{6} - {b}^{6} + 2 {a}^{3} {b}^{3} \\ \\ 9 -( ( { {a}^{3} )}^{2} +({ {b}^{3} )}^{2} - 2 {a}^{3} {b}^{3} )$
is not it seems to

${x}^{2} + {y}^{2} - 2xy = ( {x - y)}^{2}$
So

$9 - ( { {a}^{3} - {b}^{3} })^{2} \\ \\ ( {3)}^{2} - ( { {a}^{3} - {b}^{3} })^{2} \\ \\ = > since \\ \\ {x}^{2} - {y}^{2} =( x - y)(x + y) \\ \\ (3 - {a}^{3} + {b}^{3} )(3 + {a}^{3} - {b}^{3})$
are the factors of the given polynomial.