SOLUTION: Factor completely
100-c^6
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Question 606795: Factor completely
100-c^6
Found 2 solutions by richwmiller, solver91311:
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
remember that a^2-b^2=(a+b)*(a-b)
100=a^2 so a=10
b^2=c^6=(c^3)^2
(10-c^3)*(10+c^3)
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Factor the difference of two squares:
(10\ -\ c^3))
Bearing in mind that the cube root of 10 cubed is 10, factor the sum and difference of two cubes:
 \left(\sqrt[3]{100}\ -\ c\sqrt[3]{10}\ +\ c^2\right)\left(\sqrt[3]{10}\ -\ c\right)\left(\sqrt[3]{100}\ +\ c\sqrt[3]{10}\ +\ c^2\right))
You can also apply the quadratic formula to the two trinomials and get the following result:
 \left(c\ -\ \frac{c\sqrt[3]{10}\ +\ \sqrt{c^2\sqrt[3]{100}\ -\ 4\sqrt[3]{100}}}{2}\right)\left(c\ -\ \frac{c\sqrt[3]{10}\ -\ \sqrt{c^2\sqrt[3]{100}\ -\ 4\sqrt[3]{100}}}{2}\right)\cdots\\ \ \ \ \ \ \ \ \ \ \ \cdots \left(\sqrt[3]{10}\ -\ c\right) \left(c\ -\ \frac{-c\sqrt[3]{10}\ +\ \sqrt{c^2\sqrt[3]{100}\ -\ 4\sqrt[3]{100}}}{2}\right)\left(c\ -\ \frac{-c\sqrt[3]{10}\ -\ \sqrt{c^2\sqrt[3]{100}\ -\ 4\sqrt[3]{100}}}{2}\right))
Which is the end because you started with a 6th degree polynomial and ended with 6 first degree binomial factors.
John
My calculator said it, I believe it, that settles it
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