Question

2. The sum of two cubes can be factored by using the formula q3 + b3 = (a + b)(a? - ab + b).

Answer 1

Factorization involves breaking down an expression.

• The factorized expression of $8x^3 + 27$ is $8x^3 + 27 =(2x + 3)(4x^2 -8x3 +9)$
• The other factor of $2^3 - b^3$ is $4+2b+b^2$
• The factorized expression of x^3 - 1 is $x^3 - 1^3=(x-1)(x^2+x+1)$

The sum of cubes is given as:

$(a^3 + b^3) = (a + b)(a^2 -ab + b^2)$

(a) Verify the formula

Expand the expression on the right-hand side

$(a^3 + b^3) = a^3 -a^2b + ab^2 +a^2b - ab^2 + b^3$

Collect like terms

$(a^3 + b^3) = a^3 -a^2b +a^2b+ ab^2 - ab^2 + b^3$

$(a^3 + b^3) = a^3 + b^3$

The formula has been verified

(b) Factorized 8x^3+ 27

We have:

$8x^3 + 27$

Express 27 as 3^3

$8x^3 + 27 =8x^3 + 3^2$

Express 8 as 2^3

$8x^3 + 27 =2^3x^3 + 3^3$

Rewrite as:

$8x^3 + 27 =(2x)^3 + 3^3$

Given that:

$(a^3 + b^3) = (a + b)(a^2 -ab + b^2)$

The expression becomes

$8x^3 + 27 =(2x + 3)((2x)^2 - (2x)3 +3^2)$

$8x^3 + 27 =(2x + 3)(4x^2 -8x3 +9)$

(c)The other factor of 2^3 - b^3

By difference of cubes, we have:

$x^3 - y^3=(x-y)(x^2+xy+y^2)$

So, the equation becomes

$2^3 - b^3=(2-b)(2^2+2b+b^2)$

This gives

$2^3 - b^3=(2-b)(4+2b+b^2)$

Hence, the other factor of $2^3 - b^3$ is $4+2b+b^2$

(c) Factor x^3 - 1

We have:

$x^3 - y^3=(x-y)(x^2+xy+y^2)$

Express 1 as 1^3 in x^3 - 1

$x^3 - 1 =x^3 - 1^3$

$x^3 - y^3=(x-y)(x^2+xy+y^2)$ becomes

$x^3 - 1^3=(x-1)(x^2+x+1^2)$

$x^3 - 1^3=(x-1)(x^2+x+1)$

Read more about factorization at:

brainly.com/question/43919