# How do you find the zeroes of P(x)=x^6-729?

Jul 2, 2015

You may not see it right away, but $729 = {3}^{6}$
Since the exponent (6) is even it is also possible that $729 = {\left(- 3\right)}^{6}$
${x}^{6} = {3}^{6} \mathmr{and} {x}^{6} = {\left(- 3\right)}^{6} \to x = 3 \mathmr{and} x = - 3$