Use the difference of cubes for 27x^(3)-1

Answers

Answer 1

Answer:

Note:

$(a-b)^3=a^3-3a^2b+3ab^2-b^3\n\mathrm{or,\ }(a-b)^3=a^3-b^3-3ab(a-b)\n\mathrm{or,\ }a^3-b^3=(a-b)^3+3ab(a-b)\n\mathrm{or,\ }a^3-b^3=(a-b)[(a-b)^2+3ab]\n\mathrm{or,\ }a^3-b^3=(a-b)(a^2-2ab+b^2+3ab)\n\therefore\ a^3-b^3=(a-b)(a^2+ab+b^2)$

Answer:

$27x^3-1\n=(3x)^2-1\n=(3x-1)[(3x)^2+(3x)(1)+1^2]\n=(3x-1)(9x^2+3x+1)$

Answer 2

Answer:

Answer:

To factor the expression 27x^3 - 1 using the difference of cubes formula, we can follow these steps:

1. Identify the cube root of each term. In this case, the cube root of 27x^3 is 3x, and the cube root of 1 is 1.

2. Write the formula for the difference of cubes, which is: a^3 - b^3 = (a - b)(a^2 + ab + b^2).

3. Replace "a" with 3x and "b" with 1 in the formula.

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

4. Simplify the expression inside the parentheses.

(3x - 1)(9x^2 + 3x + 1)

Therefore, using the difference of cubes formula, we can factor 27x^3 - 1 as (3x - 1)(9x^2 + 3x + 1).

Step-by-step explanation:

$√(101)$

Step-by-step explanation:

The distance between two points can be found by using the formula

$d = \sqrt{ ({x1 - x2})^(2) + ({y1 - y2})^(2) } \n$

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(3,4) and (4,-6)

The distance between them is

$d = \sqrt{ ({3 - 4})^(2) + ({4 + 6})^(2) } \n = \sqrt{ ({ - 1})^(2) + {10}^(2) } \n = √(1 + 100 ) \: \: \: \: \: \: \: \: \: \n = √(101) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$