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

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

I don't really understand much especially gcf and stuff like factoring completely please help,

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Answer 1 · 2016-07-11T20:31:03

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

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

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

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

Explanation:

The difference of cubes identity can be written:

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

The sum of cubes identity can be written:

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

The difference of squares identity can be written:

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

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Question 1

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

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

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

$= (x - 2) (x^{2} + 2 x + 2^{2})$ $=(x-2)(x^2+2x+2^2)$

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

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Question 2

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

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

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

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

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

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

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

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

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

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

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Question 3

Factor by grouping:

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

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

$= x^{2} (x - 2) - 4 (x - 2)$ $=x^2(x-2)-4(x-2)$

$= (x^{2} - 4) (x - 2)$ $=(x^2-4)(x-2)$

$= (x^{2} - 2^{2}) (x - 2)$ $=(x^2-2^2)(x-2)$

$= (x - 2) (x + 2) (x - 2)$ $=(x-2)(x+2)(x-2)$

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

I don't really understand much especially gcf and stuff like factoring completely please help,

1 Answer

Explanation:

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Math

Humanities

... and beyond

My question is to please help me with factoring completly. here some that i need help on just three? 1. x3−8x^3-8 2. 276+125y327^6+125y^3 3. x3−2x2−4x+8x^3-2x^2-4x+8

I don't really understand much especially gcf and stuff like factoring completely please help,

1 Answer

Explanation:

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Impact of this question

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