How do you solve $\sqrt[3]{3 x} + 6 = 10$ $root3(3x)+6=10$ ?

Question

2110 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-14T14:03:48

$x = \frac{64}{3}$ $x=64/3$

We can take the original:

$\sqrt[3]{3 x} + 6 = 10$ $root(3)(3x)+6=10$

and subtract 6 from both sides:

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

$3 x = 64$ $3x=64$

$x = \frac{64}{3}$ $x=64/3$

Answer 2 · 2018-06-14T14:04:36

$x = \frac{64}{3}$ $x=64/3$

$\sqrt[3]{3 x} + 6 = 10$ $root3(3x)+6=10$

$\sqrt[3]{3 x} = 4$ $root3(3x)=4$

${(\sqrt[3]{3 x})}^{3} = 4^{3}$ $(root3(3x))^3=4^3$

$3 x = 64$ $3x=64$

$x = \frac{64}{3}$ $x=64/3$

How do you solve 3√3x+6=10root3(3x)+6=10?