Overview
This exercise applies two core algebraic identities for factoring sums and differences of cubes:
- (a + b)(a² − ab + b²) = a³ + b³
- (a − b)(a² + ab + b²) = a³ − b³
For a refresher on foundational algebraic manipulation, see our guide on How to Simplify Algebraic Expressions: A Step-by-Step Guide.
Worked Examples
Using (a + b)(a² − ab + b²)
Example 1: (p + 1)(p² − p + 1)
- Apply the formula with a = p, b = 1
- Result: p³ + 1³ = p³ + 1
Example 2: (3 + b)(9 − 3b + b²)
- a = 3, b = b
- Result: 3³ + b³ = 27 + b³
Example 3: (5p + 1/5p)(25p² − 1 + 1/25p²)
- a = 5p, b = 1/5p
- Result: 125p³ + 1/125p³
Using (a − b)(a² + ab + b²)
Example 4: (2 − 1)(4 + 2 + 1)
- a = 2, b = 1
- Result: 2³ − 1³ = 8 − 1 = 7
Example 5: (3p − 1/3p)(9p² + 1 + 1/9p²)
- a = 3p, b = 1/3p
- Result: 27p³ − 1/27p³
Advanced Applications
Higher Powers and Factoring
- Expressions like x⁶ − y⁶ can be factored using difference of cubes:
- x⁶ − y⁶ = (x²)³ − (y²)³ = (x² − y²)(x⁴ + x²y² + y⁴)
- Then further factor x² − y² = (x − y)(x + y)
- Similarly, x⁸ − y⁸ factors as (x⁴ − y⁴)(x⁴ + y⁴) using difference of squares
Combining Formulas
- Expressions like (2x − y)(4x² + 2xy + y²)(2x + y)(4x² − 2xy + y²)
- First factor: (2x − y)(4x² + 2xy + y²) = 8x³ − y³
- Second factor: (2x + y)(4x² − 2xy + y²) = 8x³ + y³
- Product: (8x³ − y³)(8x³ + y³) = 64x⁶ − y⁶
This builds on core polynomial concepts covered in Understanding Quadratic Polynomials: Key Concepts and Formulas.
Square Roots and Fractional Powers
- When factors involve square roots, they can cancel: (√a − √b)(√a + √b) = a − b
- Example: (√x + √y)(√x − √y) = x − y
Key Steps for Solving
- Identify the correct formula based on the sign between terms (plus or minus)
- Recognize a and b values from the given expression
- Compute a², ab, and b² for the second factor
- Apply the formula to get a³ ± b³
- Simplify by computing cubes (especially for fractions and variables)
For more practice with such techniques, review Mastering Quadratic Equations: Factoring and Quadratic Formula Explained and the broader Comprehensive Algebra 2 Guide: Binomials, Exponents, and Inverses.
If you need a refresher on the underlying properties, refer to Understanding the Distributive Property and Key Algebra Terms.
today we start exercise four point six in four point there are two formulas
which we will use in this exercise one is a plus b
whole square a square minus a b plus b square is equal to a q plus b cube and a minus b
into a square plus a b plus b square is equal to a cube
a minus b a minus b a square a square plus a b
a value is one a value is a and b value is one plus b square one whole square
it can be square a minus b a square plus a b plus b square equal to aq minus
b cube a cube minus b cube is 1 a cubed
minus 1 cube 1 this is our answer second part is a minus b into a plus
a b plus b square a minus b a formula use over a square
a is 3 3 square which is 9 plus a b 3 into b second factor
3 b plus b square b square is b square a q minus b cube 3 cube
minus b cube 3 cube 27 and a square
a square minus a b a b plus b square
a cube plus b cube a cube is equal to 512
plus b cube part 4 is a plus 2 a square minus a minus 2 a
plus 4 a plus 2 a plus b a a or e b
a square minus a b plus b square
2 square [Music] a cube and b cube
is 2 2 cube 8. question 2 part 1
a plus b a square minus a b plus b square formula convert colon a
plus b four p square is one coordinate a square
2 p square minus a b 2 p and 1 by 2 b again 2 p or 1
p whole square a plus b a square minus a b plus b square a cube
plus b cube 2 cube 8 p cube plus 1 by 8 p q
this is our required question two part three three p minus one by three p into
nine p square one cubic milligram plus a square plus a b
3p and 1 by 3 p plus b square 1 by 3 b whole square
a minus b a square plus a b plus b square a cube minus
b cube 27 p cube minus 1 by
3 power 3 27 p q part 5 5 p plus 1 by 5 p
into 25 p square one kubis molecule and one by twenty five t square a plus b
into a square 5 b whole square minus
a b plus b square 1 by 5 b whole square
a plus b a square minus a b plus b square a cube plus
b cube 125 5q pq plus 1 by 125
pq a square minus b square x square minus y square x minus y
and x plus y is my convert for that x square minus x y plus y square
x square plus x y plus y squared factory
a minus b a square plus a b plus b square raja a cube minus b cube
a plus b a square plus a b plus b square a cube plus b cube
x cube minus y cubed or x cubed plus like a cubed minus y cubed object
or formula a minus b or a plus b x6
minus y six question three part two in subco a plus b a minus b a
square minus b square yes a square x square
minus y yeah a minus b a plus b a square in a time a to a square
minus b square or easy x square power multiply
power 2 multiply 2 2 multiply 2 4 x 4 plus 5
a minus b a plus b a value x 4 or b q value y 4 a
minus b or a plus b a square in x 4 whole square minus b square y 4 whole square
x 8 minus y 8 power power multiply [Music]
plus four effectively 2 x minus y 4 x square plus 2 x y plus y square
is kasat 2x plus y casat 4 x square minus 2
x y plus y square a minus b a squared can be 2 x whole square
plus a b 2 x into y plus b square y square is a formula
a plus b a square 2 x whole square minus a b 2 x
into y n plus b square y square minus b a square plus a b plus b square
is omega a cube plus b cubed 4 power 3 8 x cube
minus y cube 8 x cube plus y q
a minus b and a plus b a a value a square minus b
square last step may 8 square 64
x 6 2 3 6 minus y six question three part four yam rearrange
x minus two plus a minus b a square plus a b x
into 2 plus b square a plus b a square minus
a b n plus b squared a cube plus
bq x cube 2 cube 8 x cube and 8. a minus b
a plus b a square minus b square x 6 minus x square 60
for question four find the product with the help of formula
is a square so under the root x cos square length
negative square root of square root cancel is a b a and b
u is k equal plus b square so formula 1 a minus b a square plus a b plus b
square a cube minus b cube
is under the root of 1 by 2 b 3 minus y 1 by 2 whole power
3 x power power multiplied with 3 or 1 by 2 multiply him with a 3 by 2 or yamp
y 3 by 2. question 5 simplify with the help of formula
xp plus y power q e by a plus b omega a square
plus a b a x power b or v y power q plus b square
y cube whole square yeah formula form a plus b a square plus a square
minus a b plus b square is say a cube minus b cube
x power 3 p minus y power 3 cube power power multiply yeah my exercise
4.6
The two core identities are: (a + b)(a² − ab + b²) = a³ + b³ for the sum of cubes, and (a − b)(a² + ab + b²) = a³ − b³ for the difference of cubes. These formulas allow you to factor expressions like a³ + b³ and a³ − b³ into a product of a binomial and a trinomial.
Identify the sign between the two terms: if the expression is a sum (e.g., a³ + b³), use (a + b)(a² − ab + b²). If it is a difference (e.g., a³ − b³), use (a − b)(a² + ab + b²). The sign in the trinomial pattern is opposite to the sign in the binomial.
Yes. For example, (5p + 1/5p)(25p² − 1 + 1/25p²) yields 125p³ + 1/125p³. For higher powers like x⁶ − y⁶, treat x² and y² as a and b respectively: (x²)³ − (y²)³ = (x² − y²)(x⁴ + x²y² + y⁴), then further factor x² − y² using difference of squares.
Factor each pair separately. For example, (2x − y)(4x² + 2xy + y²) = 8x³ − y³ and (2x + y)(4x² − 2xy + y²) = 8x³ + y³. Then multiply the results: (8x³ − y³)(8x³ + y³) = 64x⁶ − y⁶, using the difference of squares formula.
- Identify the correct formula based on the sign (plus for sum, minus for difference). 2) Recognize a and b from the given expression. 3) Compute a², ab, and b² for the second factor. 4) Apply the formula to get a³ ± b³. 5) Simplify by computing cubes, especially for fractions and variables.
When factors involve square roots, they can cancel. For example, (√a − √b)(√a + √b) = a − b (difference of squares). This technique works with sum/difference of cubes when the terms are in the form of square roots, simplifying radicals by rationalizing or combining them.
Yes. x⁸ − y⁸ can be factored as (x⁴ − y⁴)(x⁴ + y⁴) using the difference of squares. Then further factor x⁴ − y⁴ as (x² − y²)(x² + y²) and finally (x − y)(x + y). This shows how to chain multiple identities for higher-power expressions.
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