How do you factor $a^{8} - a^{2} b^{6}$ $a^8 - a^2b^6$ ?

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How do you factor $a^{8} - a^{2} b^{6}$ $a^8 - a^2b^6$ ?

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Answer 1 · 2016-05-10T14:23:17

$a^{2} (a - b) (a^{2} + a b + b^{2}) (a + b) (a^{2} - a b + b^{2})$ $a^2(a-b)(a^2+ab+b^2)(a+b)(a^2-ab+b^2)$

Explanation:

$a^{2 + 6} - a^{2} b^{6}$ $a^(2+6) - a^2b^6$

$= a^{2} a^{6} - a^{2} b^{6}$ $= a^2a^6-a^2b^6$

$= a^{2} (a^{6} - b^{6})$ $=a^2(a^6-b^6)$ , here $a^{2}$ $color(red)(a^2)$ is common between the terms

$= a^{2} (a^{3 \times 2} - b^{3 \times 2})$ $=a^2(a^(3xx2)-b^(3xx2))$

$= a^{2} [{(a^{3})}^{2} - {(b^{3})}^{2}]$ $=a^2[(a^3)^2-(b^3)^2]$

$This is of the form x^{2} - y^{2} = (x - y) (x + y), where x = a^{3} and y = b^{3}$ $color(red)("This is of the form " x^2-y^2 = (x-y)(x+y), "where " x=a^3 " and " y = b^3)$

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

Now we factorize $(a^{3} - b^{3})$ $(a^3-b^3)$ and $(a^{3} + b^{3})$ $(a^3+b^3)$ :

$We know that (a^{3} - b^{3}) = (a - b) (a^{2} + a b + b^{2})$ $color(red)("We know that " (a^3-b^3)=(a-b)(a^2+ab+b^2)$

$And (a^{3} + b^{3}) = (a + b) (a^{2} - a b + b^{2})$ $color(red)("And "(a^3+b^3)=(a+b)(a^2-ab+b^2)$

Then,

$a^{2} (a^{3} - b^{3}) (a^{3} + b^{3}) = a^{2} (a - b) (a^{2} + a b + b^{2}) (a + b) (a^{2} - a b + b^{2})$ $a^2(a^3-b^3)(a^3+b^3)=color(blue)(a^2(a-b)(a^2+ab+b^2)(a+b)(a^2-ab+b^2)$

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How do you factor $a^{8} - a^{2} b^{6}$ $a^8 - a^2b^6$ ?

1 Answer

Explanation:

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How do you factor $a^{8} - a^{2} b^{6}$ $a^8 - a^2b^6$ ?