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Basic Math Examples
m3=8m3=8
Step 1
Subtract 88 from both sides of the equation.
m3-8=0m3−8=0
Step 2
Step 2.1
Rewrite 88 as 2323.
m3-23=0m3−23=0
Step 2.2
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=ma=m and b=2b=2.
(m-2)(m2+m⋅2+22)=0(m−2)(m2+m⋅2+22)=0
Step 2.3
Simplify.
Step 2.3.1
Move 22 to the left of mm.
(m-2)(m2+2m+22)=0(m−2)(m2+2m+22)=0
Step 2.3.2
Raise 22 to the power of 22.
(m-2)(m2+2m+4)=0(m−2)(m2+2m+4)=0
(m-2)(m2+2m+4)=0(m−2)(m2+2m+4)=0
(m-2)(m2+2m+4)=0(m−2)(m2+2m+4)=0
Step 3
If any individual factor on the left side of the equation is equal to 00, the entire expression will be equal to 00.
m-2=0m−2=0
m2+2m+4=0m2+2m+4=0
Step 4
Step 4.1
Set m-2m−2 equal to 00.
m-2=0m−2=0
Step 4.2
Add 22 to both sides of the equation.
m=2m=2
m=2m=2
Step 5
Step 5.1
Set m2+2m+4m2+2m+4 equal to 00.
m2+2m+4=0m2+2m+4=0
Step 5.2
Solve m2+2m+4=0m2+2m+4=0 for mm.
Step 5.2.1
Use the quadratic formula to find the solutions.
-b±√b2-4(ac)2a−b±√b2−4(ac)2a
Step 5.2.2
Substitute the values a=1a=1, b=2b=2, and c=4c=4 into the quadratic formula and solve for mm.
-2±√22-4⋅(1⋅4)2⋅1−2±√22−4⋅(1⋅4)2⋅1
Step 5.2.3
Simplify.
Step 5.2.3.1
Simplify the numerator.
Step 5.2.3.1.1
Raise 22 to the power of 22.
m=-2±√4-4⋅1⋅42⋅1m=−2±√4−4⋅1⋅42⋅1
Step 5.2.3.1.2
Multiply -4⋅1⋅4−4⋅1⋅4.
Step 5.2.3.1.2.1
Multiply -4−4 by 11.
m=-2±√4-4⋅42⋅1m=−2±√4−4⋅42⋅1
Step 5.2.3.1.2.2
Multiply -4−4 by 44.
m=-2±√4-162⋅1m=−2±√4−162⋅1
m=-2±√4-162⋅1m=−2±√4−162⋅1
Step 5.2.3.1.3
Subtract 1616 from 44.
m=-2±√-122⋅1m=−2±√−122⋅1
Step 5.2.3.1.4
Rewrite -12−12 as -1(12)−1(12).
m=-2±√-1⋅122⋅1m=−2±√−1⋅122⋅1
Step 5.2.3.1.5
Rewrite √-1(12)√−1(12) as √-1⋅√12√−1⋅√12.
m=-2±√-1⋅√122⋅1m=−2±√−1⋅√122⋅1
Step 5.2.3.1.6
Rewrite √-1√−1 as ii.
m=-2±i⋅√122⋅1m=−2±i⋅√122⋅1
Step 5.2.3.1.7
Rewrite 1212 as 22⋅322⋅3.
Step 5.2.3.1.7.1
Factor 44 out of 1212.
m=-2±i⋅√4(3)2⋅1m=−2±i⋅√4(3)2⋅1
Step 5.2.3.1.7.2
Rewrite 4 as 22.
m=-2±i⋅√22⋅32⋅1
m=-2±i⋅√22⋅32⋅1
Step 5.2.3.1.8
Pull terms out from under the radical.
m=-2±i⋅(2√3)2⋅1
Step 5.2.3.1.9
Move 2 to the left of i.
m=-2±2i√32⋅1
m=-2±2i√32⋅1
Step 5.2.3.2
Multiply 2 by 1.
m=-2±2i√32
Step 5.2.3.3
Simplify -2±2i√32.
m=-1±i√3
m=-1±i√3
Step 5.2.4
The final answer is the combination of both solutions.
m=-1+i√3,-1-i√3
m=-1+i√3,-1-i√3
m=-1+i√3,-1-i√3
Step 6
The final solution is all the values that make (m-2)(m2+2m+4)=0 true.
m=2,-1+i√3,-1-i√3