This is just some hw question I'm stuck on
³√(x³+7) = (x+1)
and
(3x – 2)^5/6 = 2
Hello, civiliam!
Raise both sides to the $\displaystyle \frac{6}{5}$ power . . .$\displaystyle (3x - 2)^{\frac{5}{6}} \:= \:2$
. . $\displaystyle \bigg[(3x-2)^{\frac{5}{6}}\bigg]^{\frac{6}{5}} \;=\;2^{\frac{6}{5}} $
. . . . . .$\displaystyle 3x-2 \;=\;2\sqrt[5]{2} $
. . . . . . . . $\displaystyle 3x \;=\;2 + 2\sqrt[5]{2}$
. . . . . . . . .$\displaystyle x \;=\;\frac{2 + 2\sqrt[5]{2}}{3} $