How do you factor: $y = x^{6} - 16 x^{3} + 64 x$ $y= x^6 - 16x^3 +64x$ ?

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Answer 1 · 2016-04-20T14:46:23

$y = {(x - 2)}^{2} (x + 1 + i \sqrt{3}))^{2} (x + 1 - i \sqrt{3}))^{2}$ $y=(x-2)^2(x+1+isqrt 3))^2(x+1-isqrt3))^2$

y is a quadratic that is perfect square.

Factorizing this quadratic and using $(a^{3} - b^{3}) = (a - b) (a^{2} + a b + b^{2})$ $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ ,

$y = {(x^{3} - 2^{3})}^{2} = {((x - 2) (x^{2} + 2 x + 4))}^{2}$ $y=(x^3-2^3)^2=((x-2)(x^2+2x+4))^2$

$= {(x - 2)}^{2} {({(x + 1)}^{2} + {(\sqrt{3})}^{2})}^{2}$ $=(x-2)^2((x+1)^2+(sqrt 3)^2)^2$

$= {(x - 2)}^{2} (x + 1 + i \sqrt{3}))^{2} (x + 1 - i \sqrt{3}))^{2}$ $=(x-2)^2(x+1+isqrt 3))^2(x+1-isqrt3))^2$

How do you factor: y= x^6 - 16x^3 +64x y=x6−16x3+64xy= x^6 - 16x^3 +64x ?