How do you solve $\sqrt[3]{x - 6} + 6 = - 4$ $\root[ 3] { x - 6} + 6= - 4$ ?

Question

1129 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-26T20:41:56

$x = - 994$ $x=-994$

$\sqrt[3]{x - 6} + 6 = - 4$ $\root(3)(x-6)+6=-4$

Subtract $6$ $6$ from both sides:

$\sqrt[3]{x - 6} = - 10$ $\root(color(red)(3))(x-6)=-10$

Cube both sides to cancel out the cube root:

${(\sqrt[3]{x - 6})}^{3} = {(- 10)}^{3}$ $(\root(color(red)(3))(x-6))^color(red)(3)=(-10)^color(red)(3)$

Simplifying:

$x - 6 = - 1000$ $x-6=-1000$

$\to x = - 944$ $\rightarrow x=-944$

How do you solve \root[ 3] { x - 6} + 6= - 43√x−6+6=−4\root[ 3] { x - 6} + 6= - 4?