Case IV.

111. When a binomial is the difference of two cubes.

lint Since a^3 - b^3a - b = a^2 + ab + b^2,

the factors of $a^{3} - b^{3}$ are $a - b$ and $a^{2} + a b + b^{2}$ .

In like manner we can resolve into factors any expression which can be written as the difference of two cubes.

The rule for extracting the cube root of a monomial, when the monomial is a perfect cube, is,

Extract the cube root of the coefficient, and divide the index of each letter by $3$ .

Resolve into factors $8 a^{3} - 27 b^{6}$ .

Since $8 a^{3} = (2 a)^{3}$ , and $27 b^{6} = (3 b^{2})^{3}$ , we can write $8 a^{3} - 27 b^{6}$ as $(2 a)^{3} - (3 b^{2})^{3}$ .

lint Since a^3 - b^3 = (a - b)(a^2 + ab + b^2),

we have, by putting $2 a$ for $a$ and $3 b^{2}$ for $b$ ,

$(2 a)^{3} - (3 b^{2})^{3} & = (2 a - 3 b^{2}) [(2 a)^{2} + 2 a \times 3 b^{2} + (3 b^{2})^{2}] & = (2 a - 3 b^{2}) (4 a^{2} + 6 a b^{2} + 9 b^{4}) .$

Resolve into factors $64 x^{3} - 1$ .

$64 x^{3} - 1 & = (4 x)^{3} - 1 & = (4 x - 1) [(4 x)^{2} + 4 x + 1] & = (4 x - 1) (16 x^{2} + 4 x + 1) .$

find the factors of a binomial when it is the difference of two cubes, therefore,

Take the difference of the cube roots of the terms for one factor, and the sum of the squares of the cube roots of the terms plus their product for the other factor.

Exercise 35.

Resolve into factors:

$8 x^{3} - y^{3}$ .

$x^{3} - 1$ .

$x^{3} y^{3} - z^{3}$ .

$x^{3} - 64$ .

$125 a^{3} - b^{3}$ .

$a^{3} - 343$ .