# Thread: Find all Real solutions

1. ## Find all Real solutions

This is just some hw question I'm stuck on

³√(x³+7) = (x+1)

and

(3x – 2)^5/6 = 2

2. Originally Posted by civiliam
This is just some hw question I'm stuck on

³√(x³+7) = (x+1)

and

(3x – 2)^5/6 = 2
$\sqrt[3]{x^3+7}=x+1$

Cube both sides:

$x^3+7=x^3+3x^2+3x+1$

$3x^2+3x-6=0$

$x^2+x-2=0$

$(x+2)(x-1)$

$\boxed{x=-2 \ \ or \ \ x=1}$

3. Hello, civiliam!

$(3x - 2)^{\frac{5}{6}} \:= \:2$
Raise both sides to the $\frac{6}{5}$ power . . .

. . $\bigg[(3x-2)^{\frac{5}{6}}\bigg]^{\frac{6}{5}} \;=\;2^{\frac{6}{5}}$

. . . . . . $3x-2 \;=\;2\sqrt[5]{2}$

. . . . . . . . $3x \;=\;2 + 2\sqrt[5]{2}$

. . . . . . . . . $x \;=\;\frac{2 + 2\sqrt[5]{2}}{3}$

4. thank you both I think I need to review exponents