Examples
x3-8=0x3−8=0
Step 1
Add 88 to both sides of the equation.
x3=8x3=8
Step 2
Subtract 88 from both sides of the equation.
x3-8=0x3−8=0
Step 3
Step 3.1
Rewrite 88 as 2323.
x3-23=0x3−23=0
Step 3.2
Since both terms are perfect cubes, factor using the difference of cubes formula, a3-b3=(a-b)(a2+ab+b2)a3−b3=(a−b)(a2+ab+b2) where a=xa=x and b=2b=2.
(x-2)(x2+x⋅2+22)=0(x−2)(x2+x⋅2+22)=0
Step 3.3
Simplify.
Step 3.3.1
Move 22 to the left of xx.
(x-2)(x2+2x+22)=0(x−2)(x2+2x+22)=0
Step 3.3.2
Raise 22 to the power of 22.
(x-2)(x2+2x+4)=0(x−2)(x2+2x+4)=0
(x-2)(x2+2x+4)=0(x−2)(x2+2x+4)=0
(x-2)(x2+2x+4)=0(x−2)(x2+2x+4)=0
Step 4
If any individual factor on the left side of the equation is equal to 00, the entire expression will be equal to 00.
x-2=0x−2=0
x2+2x+4=0x2+2x+4=0
Step 5
Step 5.1
Set x-2x−2 equal to 00.
x-2=0x−2=0
Step 5.2
Add 22 to both sides of the equation.
x=2x=2
x=2x=2
Step 6
Step 6.1
Set x2+2x+4x2+2x+4 equal to 00.
x2+2x+4=0x2+2x+4=0
Step 6.2
Solve x2+2x+4=0x2+2x+4=0 for xx.
Step 6.2.1
Use the quadratic formula to find the solutions.
-b±√b2-4(ac)2a−b±√b2−4(ac)2a
Step 6.2.2
Substitute the values a=1a=1, b=2b=2, and c=4c=4 into the quadratic formula and solve for xx.
-2±√22-4⋅(1⋅4)2⋅1−2±√22−4⋅(1⋅4)2⋅1
Step 6.2.3
Simplify.
Step 6.2.3.1
Simplify the numerator.
Step 6.2.3.1.1
Raise 22 to the power of 22.
x=-2±√4-4⋅1⋅42⋅1x=−2±√4−4⋅1⋅42⋅1
Step 6.2.3.1.2
Multiply -4⋅1⋅4−4⋅1⋅4.
Step 6.2.3.1.2.1
Multiply -4−4 by 11.
x=-2±√4-4⋅42⋅1x=−2±√4−4⋅42⋅1
Step 6.2.3.1.2.2
Multiply -4−4 by 44.
x=-2±√4-162⋅1x=−2±√4−162⋅1
x=-2±√4-162⋅1x=−2±√4−162⋅1
Step 6.2.3.1.3
Subtract 1616 from 44.
x=-2±√-122⋅1x=−2±√−122⋅1
Step 6.2.3.1.4
Rewrite -12−12 as -1(12).
x=-2±√-1⋅122⋅1
Step 6.2.3.1.5
Rewrite √-1(12) as √-1⋅√12.
x=-2±√-1⋅√122⋅1
Step 6.2.3.1.6
Rewrite √-1 as i.
x=-2±i⋅√122⋅1
Step 6.2.3.1.7
Rewrite 12 as 22⋅3.
Step 6.2.3.1.7.1
Factor 4 out of 12.
x=-2±i⋅√4(3)2⋅1
Step 6.2.3.1.7.2
Rewrite 4 as 22.
x=-2±i⋅√22⋅32⋅1
x=-2±i⋅√22⋅32⋅1
Step 6.2.3.1.8
Pull terms out from under the radical.
x=-2±i⋅(2√3)2⋅1
Step 6.2.3.1.9
Move 2 to the left of i.
x=-2±2i√32⋅1
x=-2±2i√32⋅1
Step 6.2.3.2
Multiply 2 by 1.
x=-2±2i√32
Step 6.2.3.3
Simplify -2±2i√32.
x=-1±i√3
x=-1±i√3
Step 6.2.4
Simplify the expression to solve for the + portion of the ±.
Step 6.2.4.1
Simplify the numerator.
Step 6.2.4.1.1
Raise 2 to the power of 2.
x=-2±√4-4⋅1⋅42⋅1
Step 6.2.4.1.2
Multiply -4⋅1⋅4.
Step 6.2.4.1.2.1
Multiply -4 by 1.
x=-2±√4-4⋅42⋅1
Step 6.2.4.1.2.2
Multiply -4 by 4.
x=-2±√4-162⋅1
x=-2±√4-162⋅1
Step 6.2.4.1.3
Subtract 16 from 4.
x=-2±√-122⋅1
Step 6.2.4.1.4
Rewrite -12 as -1(12).
x=-2±√-1⋅122⋅1
Step 6.2.4.1.5
Rewrite √-1(12) as √-1⋅√12.
x=-2±√-1⋅√122⋅1
Step 6.2.4.1.6
Rewrite √-1 as i.
x=-2±i⋅√122⋅1
Step 6.2.4.1.7
Rewrite 12 as 22⋅3.
Step 6.2.4.1.7.1
Factor 4 out of 12.
x=-2±i⋅√4(3)2⋅1
Step 6.2.4.1.7.2
Rewrite 4 as 22.
x=-2±i⋅√22⋅32⋅1
x=-2±i⋅√22⋅32⋅1
Step 6.2.4.1.8
Pull terms out from under the radical.
x=-2±i⋅(2√3)2⋅1
Step 6.2.4.1.9
Move 2 to the left of i.
x=-2±2i√32⋅1
x=-2±2i√32⋅1
Step 6.2.4.2
Multiply 2 by 1.
x=-2±2i√32
Step 6.2.4.3
Simplify -2±2i√32.
x=-1±i√3
Step 6.2.4.4
Change the ± to +.
x=-1+i√3
x=-1+i√3
Step 6.2.5
Simplify the expression to solve for the - portion of the ±.
Step 6.2.5.1
Simplify the numerator.
Step 6.2.5.1.1
Raise 2 to the power of 2.
x=-2±√4-4⋅1⋅42⋅1
Step 6.2.5.1.2
Multiply -4⋅1⋅4.
Step 6.2.5.1.2.1
Multiply -4 by 1.
x=-2±√4-4⋅42⋅1
Step 6.2.5.1.2.2
Multiply -4 by 4.
x=-2±√4-162⋅1
x=-2±√4-162⋅1
Step 6.2.5.1.3
Subtract 16 from 4.
x=-2±√-122⋅1
Step 6.2.5.1.4
Rewrite -12 as -1(12).
x=-2±√-1⋅122⋅1
Step 6.2.5.1.5
Rewrite √-1(12) as √-1⋅√12.
x=-2±√-1⋅√122⋅1
Step 6.2.5.1.6
Rewrite √-1 as i.
x=-2±i⋅√122⋅1
Step 6.2.5.1.7
Rewrite 12 as 22⋅3.
Step 6.2.5.1.7.1
Factor 4 out of 12.
x=-2±i⋅√4(3)2⋅1
Step 6.2.5.1.7.2
Rewrite 4 as 22.
x=-2±i⋅√22⋅32⋅1
x=-2±i⋅√22⋅32⋅1
Step 6.2.5.1.8
Pull terms out from under the radical.
x=-2±i⋅(2√3)2⋅1
Step 6.2.5.1.9
Move 2 to the left of i.
x=-2±2i√32⋅1
x=-2±2i√32⋅1
Step 6.2.5.2
Multiply 2 by 1.
x=-2±2i√32
Step 6.2.5.3
Simplify -2±2i√32.
x=-1±i√3
Step 6.2.5.4
Change the ± to -.
x=-1-i√3
x=-1-i√3
Step 6.2.6
The final answer is the combination of both solutions.
x=-1+i√3,-1-i√3
x=-1+i√3,-1-i√3
x=-1+i√3,-1-i√3
Step 7
The final solution is all the values that make (x-2)(x2+2x+4)=0 true.
x=2,-1+i√3,-1-i√3