Worked example 16: Factorising a difference of two cubes
QUESTION
Factorise: 16 y^3 432.
SOLUTION
Step 1: Take out the common factor 16
16 y^3 432 = 16
(
y^3 27
)
Step 2: Take the cube root of terms that are perfect cubes
We are working with the difference of two cubes. We know thatx^3 y^3 = (x y)
(
x^2 +xy+y^2
)
, so we
need to identifyxandy.
We start by noting that^3
√
y^3 =yand^3
p
27 = 3. These give the terms in the first bracket. This also tells us that
x=yandy= 3.
Step 3: Find the three terms in the second bracket
We can replacexandyin the factorised form of the expression for the difference of two cubes withyand 3.
Doing so we get the second bracket:
16
(
y^3 27
)
= 16 (y 3)
(
y^2 + 3y+ 9
)
Step 4: Expand the brackets to check that the expression has been correctly factorised
16(y 3)(y^2 + 3y+ 9) = 16[(y(y^2 + 3y+ 9) 3(y^2 + 3y+ 9)]
= 16[y^3 + 3y^2 + 9y 3 y^2 9 y 27]
= 16y^3 432
Worked example 17: Factorising a sum of two cubes
QUESTION
Factorise: 8 t^3 + 125p^3.
SOLUTION
Step 1: Take the cube root of terms that are perfect cubes
We are working with the sum of two cubes. We know thatx^3 +y^3 = (x+y)
(
x^2 xy+y^2
)
, so we need to
identifyxandy.
We start by noting that
p 3
8 t^3 = 2tand^3
√
125 p^3 = 5p. These give the terms in the first bracket. This also tells
us thatx= 2tandy= 5p.
Chapter 1. Algebraic expressions 29