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Factoring Perfect Cubes
Perfect Cubes:
The cube of a number is that number raised to the third power. The number that results is called a
perfect cube.
Example:
23 = 2 × 2 × 2 = 8
The cube of 2 is 8 and 8 is the perfect cube of 2.
33 = 3 × 3 × 3 = 27
The cube of 3 is 27 and 27 is the perfect cube of3.
(−4) 3 = (−4)(−4)(−4) = −64 The cube of -4 is -64 and -64 is the perfect cube of4.
( x) 3 = ( x)( x)( x) = x 3
The cube of x is x 3 and x 3 is the perfect cube of x.
( x 2 ) 3 = ( x 2 )( x 2 )( x 2 ) = x 6
The cube of x 2 is x 6 and x 6 is the perfect cube of x 2 .
It will be very handy to know the perfect cubes of some numbers, especially the numbers one
through six.
Example: Determine the perfect cubes of the numbers one through six.
Solution: The following are the perfect cubes of these numbers.
13 = 1
23 = 8
33 = 27
4 3 = 64
53 = 125
6 3 = 216
In addition, any variable raised to power that is a multiple of 3 is a perfect cube. The following
expressions are all perfect cubes.
x9 = (x3 )3
x9
y 12
x15 y 3
x 6 y 30 z 18
Each is the result of raising the variables to the 3rd power
y 12 = ( y 4 ) 3
x 15 y 3 = ( x 5 y ) 3
x 6 y 30 z 18 ( x 2 y 10 z 6 ) 3
Factoring the Sum or Difference of Two Perfect Cubes:
To factor the sum or difference of two perfect cubes we will use the following formula.
a 3 ± b 3 = (a ± b)(a 2 µ ab + b 2 )
To use this formula you must first recognize that a binomial is either the sum or difference of
two perfect cubes. Then, apply the formula as demonstrated in the following examples. The sign
in the factored binomial is always the same as the sign in the original binomial and the sign in
front of the second term of the trinomial is always the opposite of the sign in the original
binomial. That is what the signs ( ± ) and ( µ ) mean in the formula. The last term of the trinomial is
always positive.
Example: Factor the binomial x 3 + 8 .
Solution: Since x 3 + 8 = ( x) 3 + (2) 3 both terms are perfect cubes and we may apply the formula,
a 3 ± b 3 = (a ± b)(a 2 µ ab + b 2 )
where a = x, and b = 2
x 3 + 8 = ( x + 2)( x 2 − 2 x + 4)
Be careful with the signs. The sign in the factored binomial is always the same as the sign in the
original binomial and the sign in front of the second term of the trinomial is always the opposite
of the sign in the original binomial. That is what the signs ( ± ) and ( µ) mean in the formula. The
last term of the trinomial is always positive.
Example: Factor the binomial p 3 − 64 .
Solution: Since p 3 − 64 = ( p ) 3 − (4) 3 both terms are perfect cubes and we may apply the
formula,
a 3 ± b 3 = (a ± b)(a 2 µ ab + b 2 )
where a = p, and b = 4
p 3 − 64 = ( p − 4)( p 2 + 4 p + 16)
Once again, be careful with the signs. The sign in the factored binomial is always the same as the
sign in the original binomial and the sign in front of the second term of the trinomial is always
the opposite of the sign in the original binomial. That is what the signs ( ± ) and ( µ) mean. The last
term of the trinomial is always positive.
Example: Factor the binomial 27 x 6 + 125 y 3 .
Solution: Since 27 x 6 + 125 y 3 = (3 x 2 ) 3 + (5 y ) 3 both terms are perfect cubes and we may apply
the formula,
a 3 ± b 3 = (a ± b)(a 2 µ ab + b 2 )
where a = 3x 2 , and b = 5 y
27 x 6 + 125 y 3 = (3 x 2 + 5 y )(9 x 4 − 15 x 2 y + 25 y 2 )
Example: Factor the binomial 8 p 6 q 9 − r 12 s 15 .
Solution: Since 8 p 6 q 9 − r 12 s 15 = (2 p 2 q 3 ) 3 − (r 4 s 5 ) 3 both terms are perfect cubes and we may
apply the formula,
a 3 ± b 3 = (a ± b)(a 2 µ ab + b 2 )
Where a = 2 p 2 q 3 , and b = r 4 s 5
8 p 6 q 9 − r 12 s 15 = (2 p 2 q 3 − r 4 s 5 )(4 p 4 q 6 + 2 p 2 q 3 r 4 s 5 + r 8 s 10 )
Example: Factor the binomial 5 x 3 − 40 .
Solution: First, factor out the common factor of 5 to obtain 5( x 3 − 8) . Then, since
x 3 − 8 = ( x) 3 − (2) 3 both terms are perfect cubes and we may apply the formula,
a 3 ± b 3 = (a ± b)(a 2 µ ab + b 2 )
where a = x, and b = 2
5( x 3 − 8) = 5( x − 2)( x 2 + 2 x + 4)
Example: Factor the binomial p 5 + 64 p 2 .
Solution: First, factor out the common factor of p 2 to obtain p 2 ( p 3 + 64) . Then, since
p 3 + 64 = ( p ) 3 + (4) 3 both terms are perfect cubes and we may apply the formula,
a 3 ± b 3 = (a ± b)(a 2 µ ab + b 2 )
Where a = p, and b = 4
p 2 ( p 3 + 64) = ( p + 4)( p 2 − 4 p + 16)
Example: Factor the binomial ( x + 1) 3 + ( x − 1) 3 .
Solution: Since both terms are perfect cubes we may apply the formula,
a 3 ± b 3 = (a ± b)(a 2 µ ab + b 2 )
where a = (x+1), and b =(x-1)
( x + 1) 3 + ( x − 1) 3 =
[( x + 1) + ( x − 1)][( x + 1) 2 − ( x + 1)( x − 1) + ( x − 1) 2 ]
[ 2 x ][( x 2 + 2 x + 1) − ( x 2 − 1) + ( x 2 − 2 x + 1)]
[ 2 x ][ x 2 + 3]
Applications:
Example: If the radius of a spherical balloon increases from r1 to r2 then the increase in the
volume of the balloon is given by the expression
4 3 4 3
πr2 − πr1
3
3
Write the expression in factored form.
Solution: Start by factoring out the common factor of
4
π , then factor as the difference of two
3
perfect cubes.
4 3 4 3
πr2 − πr1
3
3
4
π (r23 − r13 )
3
4
π (r2 − r1 )(r22 + r1 r2 + r12 )
3
Example: A furniture store sells two sizes of cubed shaped stackable storage containers. The
dimensions of the smaller stackable cubes is x by x by x. The dimensions of the larger stackable
cubes is y by y by y. Write an expression in factored form that represents the combined storage
space of one large and one small cube.
Solution: The volume of the smaller cube is V = x 3 . And the volume of the larger cube is
V = y 3 . To find the combined storage space (volume) we must add these expressions together to
obtain
x3 + y3
Factor this expression as a sum of two perfect cubes.
x 3 + y 3 = ( x + y )( x 2 − xy + y 2 )