How do you factor $- 49 + y^{6}$ $-49 + y^6$ ?

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Answer 1 · 2015-05-31T00:02:17

$y^{6} - 49 = (y^{3} - 7) (y^{3} + 7)$ $y^6 - 49 = (y^3 - 7)(y^3 + 7)$

Answer 2 · 2015-05-31T00:31:53

$y^{6} - 49 = (y^{3} + 7) (y^{3} - 7)$ $y^6-49=(y^3+7)(y^3-7)$

Problem: Factor $- 49 + y^{6}$ $-49+y^6$ .

Rewrite the equation as $y^{6} - 49$ $y^6-49$ .

This is an example of the difference of squares:

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

$a = y^{3}$ $a=y^3$
$b = 7$ $b=7$

${(y^{3})}^{2} - 7^{2} = (y^{3} + 7) (y^{3} - 7)$ $(y^3)^2-7^2=(y^3+7)(y^3-7)$

Answer 3 · 2015-05-31T09:04:24

$- 49 + y^{6} = y^{6} - 49 = {(y^{3})}^{2} - 7^{2} = (y^{3} - 7) (y^{3} + 7)$ $-49+y^6 = y^6-49 = (y^3)^2-7^2 = (y^3 - 7)(y^3 + 7)$

This is as far as you can go with rational coefficients.

If you are allowed real coefficients:

$y^{3} - 7 = y^{3} - {(\sqrt[3]{7})}^{3}$ $y^3-7 = y^3-(root(3)(7))^3$

$= (y - \sqrt[3]{7}) (y^{2} + (\sqrt[3]{7}) y + {(\sqrt[3]{7})}^{2})$ $= (y - root(3)(7))(y^2+(root(3)(7))y+(root(3)(7))^2)$

and

$y^{3} + 7 = y^{3} + {(\sqrt[3]{7})}^{3}$ $y^3+7 = y^3+(root(3)(7))^3$

$= (y + \sqrt[3]{7}) (y^{2} - (\sqrt[3]{7}) y + {(\sqrt[3]{7})}^{2})$ $= (y + root(3)(7))(y^2-(root(3)(7))y+(root(3)(7))^2)$

For shorthand I will write $ρ = \sqrt[3]{7}$ $rho = root(3)(7)$

$- 49 + y^{6}$ $-49+y^6$

$= (y - ρ) (y^{2} + ρ y + ρ^{2}) (y + ρ) (y^{2} - ρ y + ρ^{2})$ $= (y - rho)(y^2+rho y+rho^2)(y + rho)(y^2-rho y+rho^2)$

How do you factor −49+y6-49 + y^6?