Case V.

113. When a binomial is the sum 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 $a^2-ab+b^2$.

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

1. Resolve into factors $8x^3+27y^3$.

Since by § 112, $8x^3=(2x{)}^3$ and $27y^3=(3y{)}^3$, we can write $8x^3+27y^3$ as $(2x{)}^3+(3y{)}^3$.

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

we have, by putting $2x$ for $a$, and $3y$ for $b$,

$$\begin{gathered} (2x{)}^3+(3y{)}^3\&=(2x+3y)[(2x{)}^2-2x\times 3y+(3y{)}^2] \\ \&=(2x+3y)(4x^2-6xy+9y^2). \end{gathered}$$

2. Resolve into factors $125a^3+64x^6$.

$125a^3=(5a{)}^3,64x^6=(4x^2{)}^3;$

$$\begin{gathered} \therefore 125a^3+64x^6\&=(5a+4x^2)[(5a{)}^2-5a\times 4x^2+(4x^2{)}^2] \\ \&=(5a+4x^2)(25a^2-20a{x}^2+16x^4) \end{gathered}$$

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

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

Exercise 36.

Resolve into factors:

1. $x^3+1$.

2. $8x^3+y^3$.

3. $x^3+125$.

4. $64a^3+27$.

5. $x^3{y}^3+z^3$.

6. $a^3+64$.

7. $8a^6+b^3$.