Tutorial on how to solve equations containing cube roots. Detailed solutions to examples, explanations and exercises are included.

The idea behind solving equations containing cube roots is to raise to power 3 in order to clear the cube root using the property

- Rewrite equation with the term containing cube root on one side as follows.

^{3}√x = x

- Raise both sides to power 3 in order to clear the cube root.

(^{3}√x )^{ 3}= x^{ 3}

- Rewrite the above equation with right side equal to zero.

x - x^{ 3}= 0

- Factor

x (1 - x^{ 2}) = 0

- and solve for x.

solutions are : x = 0 , x = - 1 and x = 1.

1. x = 0__Check the solutions found.__

Left side (LS) of the given equation when x = 0

LS =^{3}√x - x =^{3}√(0) - 0 = 0

Right Side (RS) of the given equation when x = 0

RS = 0

2. x = -1

Left side (LS) of the given equation when x = -1

LS =^{3}√x - x =^{3}√(-1) - (-1) = -1 + 1 = 0

Right Side (RS) of the given equation when x = -1

RS = 0

3. x = 1

Left side (LS) of the given equation when x = 1

LS =^{3}√x - x =^{3}√(1) - 1 = 0

Right Side (RS) of the given equation when x = 1

RS = 0

- Given

^{3}√( x^{ 2}+ 2 x + 8 ) = 2

- We raise both sides to power 3 in order to clear the cube root.

[^{3}√( x^{ 2}+ 2 x + 8 ) ]^{ 3}= 2^{ 3}

- and simplify.

x^{ 2}+ 2 x + 8 = 8

- Rewrite the above equation with right side equal to zero.

x^{ 2}+ 2 x = 0

- Factor

x (x + 2) = 0

- and solve for x.

x = 0 and x = - 2.

Let us check the solutions obtained as an exercise.

1. x = 0

Left side (LS) of the given equation when x = 0

LS =^{3}√( x^{ 2}+ 2 x + 8 ) = cube_root (0 + 0 + 8) = 2

Right Side (RS) of the given equation when x = 0

RS = 2

2. x = -2

Left side (LS) of the given equation when x = 0

LS =^{3}√( x^{ 2}+ 2 x + 8 )

=^{3}√( (-2)^{ 2}+ 2*(-2) + 8 ) = cube_root ( 8 ) = 2

Right Side (RS) of the given equation when x = 0

RS = 2

Solve the following equations

1.

2.

__Solutions to above exercises__

1. x = 0 , x = 1 / 8 , x = - 1 / 8

2. x = 1 , x = -3