Factor (3x^4-3x)-(3x-3x^4)

$(3 x^{4} - 3 x) - (3 x - 3 x^{4})$

Step 1

Apply the distributive property.

$3 x^{4} - 3 x - (3 x) - (- 3 x^{4})$

Step 2

Multiply $3$ by $- 1$ .

$3 x^{4} - 3 x - 3 x - (- 3 x^{4})$

Step 3

Multiply $- 3$ by $- 1$ .

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

Step 4

Add $3 x^{4}$ and $3 x^{4}$ .

$6 x^{4} - 3 x - 3 x$

Step 5

Subtract $3 x$ from $- 3 x$ .

$6 x^{4} - 6 x$

Step 6

Rewrite $6 x^{4} - 6 x$ in a factored form.

Factor $6 x$ out of $6 x^{4} - 6 x$ .

Factor $6 x$ out of $6 x^{4}$ .

$6 x (x^{3}) - 6 x$

Factor $6 x$ out of $- 6 x$ .

$6 x (x^{3}) + 6 x (- 1)$

Factor $6 x$ out of $6 x (x^{3}) + 6 x (- 1)$ .

$6 x (x^{3} - 1)$

Rewrite $1$ as $1^{3}$ .

$6 x (x^{3} - 1^{3})$

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = x$ and $b = 1$ .

$6 x ((x - 1) (x^{2} + x \cdot 1 + 1^{2}))$

Factor.

Simplify.

Multiply $x$ by $1$ .

$6 x ((x - 1) (x^{2} + x + 1^{2}))$

One to any power is one.

$6 x ((x - 1) (x^{2} + x + 1))$

Remove unnecessary parentheses.

$6 x (x - 1) (x^{2} + x + 1)$

Factor (3x^4-3x)-(3x-3x^4)

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