# My question is to please help me with factoring completly. here some that i need help on just three? 1. x^3-8 2. 27^6+125y^3 3. x^3-2x^2-4x+8

Jul 11, 2016

1) ${x}^{3} - 8 = \left(x - 2\right) \left({x}^{2} + 2 x + 4\right)$

2) ${27}^{6} + 125 {y}^{3} = \left(729 + 5 y\right) \left(531441 - 3645 y + 25 {y}^{2}\right)$

$27 {x}^{6} + 125 {y}^{3} = \left(3 {x}^{2} + 5 y\right) \left(9 {x}^{4} - 15 {x}^{2} y + 25 {y}^{2}\right)$

3) ${x}^{3} - 2 {x}^{2} - 4 x + 8 = \left(x - 2\right) \left(x + 2\right) \left(x - 2\right)$

#### Explanation:

The difference of cubes identity can be written:

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

The sum of cubes identity can be written:

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

$\textcolor{w h i t e}{}$
Question 1

Use the difference of cubes identity with $a = x$ and $b = 2$:

${x}^{3} - 8$

$= {x}^{3} - {2}^{3}$

$= \left(x - 2\right) \left({x}^{2} + 2 x + {2}^{2}\right)$

$= \left(x - 2\right) \left({x}^{2} + 2 x + 4\right)$

$\textcolor{w h i t e}{}$
Question 2

Use the sum of cubes identity with $a = {27}^{2} = 729$ and $b = 5 y$:

${27}^{6} + 125 {y}^{3}$

$= {\left({27}^{2}\right)}^{3} + {\left(5 y\right)}^{3}$

$= \left(729 + 5 y\right) \left({729}^{2} - 729 \cdot 5 y + {\left(5 y\right)}^{2}\right)$

$= \left(729 + 5 y\right) \left(531441 - 3645 y + 25 {y}^{2}\right)$

Actually I suspect a typo in the question. If ${27}^{6}$ was supposed to be $27 {x}^{6}$, then use the sum of cubes identity with $a = 3 {x}^{2}$ and $b = 5 y$ to find:

$27 {x}^{6} + 125 {y}^{3}$

$= {\left(3 {x}^{2}\right)}^{3} + {\left(5 y\right)}^{3}$

$= \left(3 {x}^{2} + 5 y\right) \left({\left(3 {x}^{2}\right)}^{2} - \left(3 {x}^{2}\right) \left(5 y\right) + {\left(5 y\right)}^{2}\right)$

$= \left(3 {x}^{2} + 5 y\right) \left(9 {x}^{4} - 15 {x}^{2} y + 25 {y}^{2}\right)$

$\textcolor{w h i t e}{}$
Question 3

Factor by grouping:

${x}^{3} - 2 {x}^{2} - 4 x + 8$

$= \left({x}^{3} - 2 {x}^{2}\right) - \left(4 x - 8\right)$

$= {x}^{2} \left(x - 2\right) - 4 \left(x - 2\right)$

$= \left({x}^{2} - 4\right) \left(x - 2\right)$

$= \left({x}^{2} - {2}^{2}\right) \left(x - 2\right)$

$= \left(x - 2\right) \left(x + 2\right) \left(x - 2\right)$