Subtract 31 from both sides of the equation.

$\displaystyle \sqrt{x-9}+31-31 = 35-31$

$\displaystyle \sqrt{x-9}=4$

Square both sides.

$\displaystyle (\sqrt{x-9})^2=4^2$

Simplify both sides.

$\displaystyle x-9 =16$

Add nine on both sides.

$\displaystyle x-9 +9=16+9$

The answer is:  $\displaystyle 25$

To eliminate the radical, we will have to cube both sides.

$\displaystyle (\sqrt[3]{3x-7})^3=4^3$

The equation becomes:

$\displaystyle 3x-7=64$

Add seven on both sides.

$\displaystyle 3x=71$

Divide by three on both sides.

$\displaystyle \frac{3x}{3}=\frac{71}{3}$

The answer is:  $\displaystyle \frac{71}{3}$

Divide by three on both sides.

$\displaystyle \frac{3\sqrt[4]{9-2x} }{3}= \frac{1}{3}$

Rewrite the equation.

$\displaystyle \sqrt[4]{9-2x} = \frac{1}{3}$

Raise both sides to the fourth power.

$\displaystyle (\sqrt[4]{9-2x})^4 =( \frac{1}{3})^4$

Simplify both sides.

$\displaystyle 9-2x = \frac{1}{81}$

Multiply by 81 on both sides to eliminate the fraction.

$\displaystyle 81(9-2x )= \frac{1}{81} \cdot 81$

$\displaystyle 729-162x=1$

Subtract 729 from both sides.

$\displaystyle 729-162x-729=1-729$

$\displaystyle -162x = -728$

Divide by negative 162 on both sides.

$\displaystyle \frac{-162x}{-162} =\frac{ -728}{-162}$

Reduce the fractions.  Double negatives will negate, turning the answer positive.

$\displaystyle \frac{ -728}{-162} =\frac{364\times 2}{81\times 2}$

The answer is:  $\displaystyle \frac{364}{81}$

Isolate the radical.  Add four on both sides.

$\displaystyle 3\sqrt[4]{4-2x}-4+4=2+4$

$\displaystyle 3\sqrt[4]{4-2x}=6$

Divide by three on both sides.  The equation becomes:

$\displaystyle \sqrt[4]{4-2x}= 2$

Raise both sides by four.

$\displaystyle (\sqrt[4]{4-2x}) ^4= 2^4$

$\displaystyle 4-2x = 16$

Subtract four on both sides.

$\displaystyle 4-2x -4= 16-4$

$\displaystyle -2x=12$

Divide by negative two on both sides.

$\displaystyle \frac{-2x}{-2}=\frac{12}{-2}$

The answer is:  $\displaystyle -6$

Multiply by six on both sides of the equation.

$\displaystyle \frac{\sqrt[3]{5x-2}}{6} \cdot 6= \frac{1}{2} \cdot 6$

The equation becomes:

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

Cube both sides to eliminate the radical.

$\displaystyle (\sqrt[3]{5x-2}) ^3=3^3$

$\displaystyle 5x-2 = 27$

Add two on both sides.

$\displaystyle 5x-2 +2= 27+2$

$\displaystyle 5x=29$

Divide by five on both sides.

$\displaystyle \frac{5x}{5}=\frac{29}{5}$

The answer is:  $\displaystyle \frac{29}{5}$

Add seven on both sides.

$\displaystyle \sqrt{-6x-5}-7 +7= 10+7$

$\displaystyle \sqrt{-6x-5} = 17$

Square both sides.  This will eliminate the radical.

$\displaystyle (\sqrt{-6x-5})^2= 17^2$

$\displaystyle -6x-5 = 289$

Add five on both sides.

$\displaystyle -6x-5 +5= 289+5$

$\displaystyle -6x=294$

Divide both sides by negative 6 and reduce the fraction.

$\displaystyle \frac{-6x}{-6}=\frac{294}{-6}$

The answer is:  $\displaystyle -49$

Cube both sides of the equation.

$\displaystyle (\sqrt[3]{3x}) ^3= (\frac{2}{3})^3$

This will eliminate the radical on the left side.

$\displaystyle 3x= (\frac{2}{3})(\frac{2}{3})(\frac{2}{3})$

$\displaystyle 3x= \frac{8}{27}$

Divide by three on both sides.  This is similar to multiplying one-third on both sides.

$\displaystyle 3x \times \frac{1}{3}= \frac{8}{27} \times \frac{1}{3}=\frac{8}{81}$

The answer is:  $\displaystyle \frac{8}{81}$

Subtract both sides by three.

$\displaystyle 3-\sqrt[3]{2-x} -3= -1-3$

The equation becomes: 

$\displaystyle -\sqrt[3]{2-x} =-4$

Divide both sides by negative one.

$\displaystyle \frac{-\sqrt[3]{2-x} }{-1}=\frac{-4}{-1}$

$\displaystyle \sqrt[3]{2-x} =4$

Cube both sides to eliminate the radical.

$\displaystyle (\sqrt[3]{2-x}) ^3=4^3$

$\displaystyle 2-x = 64$

Add $\displaystyle x$ on both sides.

$\displaystyle 2-x +x= 64+x$

$\displaystyle 2= 64+x$

Subtract 64 on both sides.

$\displaystyle 2-64= 64+x-64$

The answer is:  $\displaystyle -62$

Raise both sides by the power of eight to eliminate the radical.

$\displaystyle (\sqrt[8]{2x-1} )^8=2^8$

Simplify both sides.

$\displaystyle 2x-1=256$

Add one on both sides.

$\displaystyle 2x-1+1=256+1$

$\displaystyle 2x=257$

Divide by two on both sides.

$\displaystyle \frac{2x}{2}=\frac{257}{2}$

The answer is:  $\displaystyle \frac{257}{2}$

Multiply by $\displaystyle \sqrt{3x}$ on both sides.

$\displaystyle \frac{1}{\sqrt{3x}} \cdot \sqrt{3x}= 4 \cdot \sqrt{3x}$

$\displaystyle 1=4\sqrt{3x}$

Divide by 4 on both sides.

$\displaystyle \frac{1}{4}=\frac{4\sqrt{3x}}{4}$

$\displaystyle \frac{1}{4} =\sqrt{3x}$

Square both sides to eliminate the radical.

$\displaystyle (\frac{1}{4}) ^2=(\sqrt{3x})^2$

$\displaystyle \frac{1}{16} = 3x$

Divide by three on both sides.  This is similar to multiplying both sides by one third.

$\displaystyle \frac{1}{16} \cdot \frac{1}{3}= 3x\cdot \frac{1}{3}$

The answer is:  $\displaystyle \frac{1}{48}$