How do you factor $64 - c^{12}$ $64 - c^12$ ?

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How do you factor $64 - c^{12}$ $64 - c^12$ ?

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Answer 1 · 2015-12-15T00:13:55

Use some special identities to find:

$64 - c^{12} = (\sqrt{2} - c) (\sqrt{2} + c) (2 + c^{2}) (2 + \sqrt{2} c + c^{2}) (2 - \sqrt{2} c + c^{2}) (2 + \sqrt{6} c + c^{2}) (2 - \sqrt{6} c + c^{2})$ $64-c^12=(sqrt(2)-c)(sqrt(2)+c)(2+c^2)(2+sqrt(2)c+c^2)(2-sqrt(2)c+c^2)(2+sqrt(6)c+c^2)(2-sqrt(6)c+c^2)$

Explanation:

The difference of cubes identity can be written:

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

The difference of squares identity can be written:

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

Note also that:

$(a^{2} + a b + b^{2}) (a^{2} - a b + b^{2}) = a^{4} + a^{2} b^{2} + b^{4}$ $(a^2+ab+b^2)(a^2-ab+b^2) = a^4+a^2b^2+b^4$

The final identity we will use is:

$(a^{2} + \sqrt{3} a b + b^{2}) (a^{2} - \sqrt{3} a b + b^{2}) = a^{4} - a^{2} b^{2} + b^{4}$ $(a^2+sqrt(3)ab+b^2)(a^2-sqrt(3)ab+b^2) = a^4-a^2b^2+b^4$

These last two identities were found by looking at:

$(a^{2} + k a b + b^{2}) (a^{2} - k a b + b^{2}) = a^{4} + (2 - k^{2}) a^{2} b^{2} + b^{4}$ $(a^2+kab+b^2)(a^2-kab+b^2) = a^4+(2-k^2)a^2b^2+b^4$

and picking $k = 1$ $k = 1$ or $k = \sqrt{3}$ $k = sqrt(3)$ to get the required result.

Note that $64 = 4^{3}$ $64 = 4^3$ and $c^{12} = {(c^{4})}^{3}$ $c^12 = (c^4)^3$ are both perfect cubes. So we can use the difference of cubes identity to start to factorise this.

Letting $a = 4$ $a=4$ and $b = c^{4}$ $b=c^4$ we find:

$64 - c^{12}$ $64-c^12$

$= 4^{3} - {(c^{4})}^{3}$ $=4^3-(c^4)^3$

$= (4 - c^{4}) (4^{2} + 4 c^{4} + {(c^{4})}^{2})$ $=(4-c^4)(4^2+4c^4+(c^4)^2)$

$= (4 - c^{4}) (16 + 4 c^{4} + c^{8})$ $=(4-c^4)(16+4c^4+c^8)$

Next both $4 = 2^{2}$ $4=2^2$ and $c^{4} = {(c^{2})}^{2}$ $c^4 = (c^2)^2$ are perfect squares so we can factor some more:

$= (2^{2} - {(c^{2})}^{2}) (16 + 4 c^{4} + c^{8})$ $=(2^2-(c^2)^2)(16+4c^4+c^8)$

$= (2 - c^{2}) (2 + c^{2}) (16 + 4 c^{4} + c^{8})$ $=(2-c^2)(2+c^2)(16+4c^4+c^8)$

Next use the difference of squares identity to factor the first quadratic:

$= ({\sqrt{2}}^{2} - c^{2}) (2 + c^{2}) (16 + 4 c^{4} + c^{8})$ $=(sqrt(2)^2-c^2)(2+c^2)(16+4c^4+c^8)$

$= (\sqrt{2} - c) (\sqrt{2} + c) (2 + c^{2}) (16 + 4 c^{4} + c^{8})$ $=(sqrt(2)-c)(sqrt(2)+c)(2+c^2)(16+4c^4+c^8)$

Next use our special $a^{4} + a^{2} b^{2} + b^{4}$ $a^4+a^2b^2+b^4$ identity with $a = 2$ $a=2$ and $b = c^{2}$ $b=c^2$ to factor some more:

$= (\sqrt{2} - c) (\sqrt{2} + c) (2 + c^{2}) (4 + 2 c^{2} + c^{4}) (4 - 2 c^{2} + c^{4})$ $=(sqrt(2)-c)(sqrt(2)+c)(2+c^2)(4+2c^2+c^4)(4-2c^2+c^4)$

Then use it again with $a = \sqrt{2}$ $a=sqrt(2)$ and $b = c$ $b=c$ to factor some more:

$= (\sqrt{2} - c) (\sqrt{2} + c) (2 + c^{2}) (2 + \sqrt{2} c + c^{2}) (2 - \sqrt{2} c + c^{2}) (4 - 2 c^{2} + c^{4})$ $=(sqrt(2)-c)(sqrt(2)+c)(2+c^2)(2+sqrt(2)c+c^2)(2-sqrt(2)c+c^2)(4-2c^2+c^4)$

Then use our final identity with $a = \sqrt{2}$ $a=sqrt(2)$ and $b = c$ $b=c$ to factor the last quartic:

$= (\sqrt{2} - c) (\sqrt{2} + c) (2 + c^{2}) (2 + \sqrt{2} c + c^{2}) (2 - \sqrt{2} c + c^{2}) (2 + \sqrt{6} c + c^{2}) (2 - \sqrt{6} c + c^{2})$ $=(sqrt(2)-c)(sqrt(2)+c)(2+c^2)(2+sqrt(2)c+c^2)(2-sqrt(2)c+c^2)(2+sqrt(6)c+c^2)(2-sqrt(6)c+c^2)$

That's as far as we can get with Real coefficients.

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