What are all the real zeros of $y = {(x - 12)}^{3} - 10$ $y=(x-12)^3-10$ ?

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Answer 1 · 2017-08-20T12:42:24

The real zero is:

$x = 12 + \sqrt[3]{10}$ $x = 12+root(3)(10)$

Given:

$y = {(x - 12)}^{3} - 10$ $y = (x-12)^3-10$

The real zero is:

$x = 12 + \sqrt[3]{10}$ $x = 12+root(3)(10)$

The other zeros are both non-real complex:

$x = 12 + ω \sqrt[3]{10}$ $x = 12+omega root(3)(10)$

$x = 12 + ω^{2} \sqrt[3]{10}$ $x = 12+omega^2 root(3)(10)$

where $ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ $omega = -1/2+sqrt(3)/2i$ is the primitive complex cube root of $1$ $1$ .

What are all the real zeros of y=(x−12)3−10y=(x-12)^3-10?