Algebra Examples

$a^{3} x^{2} - 16 a^{3} x + 64 a^{3} - b^{3} x^{2} + 16 b^{3} x - 64 b^{3}$

Step 1

Regroup terms.

$a^{3} x^{2} - 16 a^{3} x + 64 a^{3} - b^{3} x^{2} + 16 b^{3} x - 64 b^{3}$

Step 2

Factor $a^{3}$ out of $a^{3} x^{2} - 16 a^{3} x + 64 a^{3}$ .

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Step 2.1

Factor $a^{3}$ out of $a^{3} x^{2}$ .

$a^{3} (x^{2}) - 16 a^{3} x + 64 a^{3} - b^{3} x^{2} + 16 b^{3} x - 64 b^{3}$

Step 2.2

Factor $a^{3}$ out of $- 16 a^{3} x$ .

$a^{3} (x^{2}) + a^{3} (- 16 x) + 64 a^{3} - b^{3} x^{2} + 16 b^{3} x - 64 b^{3}$

Step 2.3

Factor $a^{3}$ out of $64 a^{3}$ .

$a^{3} (x^{2}) + a^{3} (- 16 x) + a^{3} \cdot 64 - b^{3} x^{2} + 16 b^{3} x - 64 b^{3}$

Step 2.4

Factor $a^{3}$ out of $a^{3} (x^{2}) + a^{3} (- 16 x)$ .

$a^{3} (x^{2} - 16 x) + a^{3} \cdot 64 - b^{3} x^{2} + 16 b^{3} x - 64 b^{3}$

Step 2.5

Factor $a^{3}$ out of $a^{3} (x^{2} - 16 x) + a^{3} \cdot 64$ .

$a^{3} (x^{2} - 16 x + 64) - b^{3} x^{2} + 16 b^{3} x - 64 b^{3}$

Step 3

Factor using the perfect square rule.

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Step 3.1

Rewrite $64$ as $8^{2}$ .

$a^{3} (x^{2} - 16 x + 8^{2}) - b^{3} x^{2} + 16 b^{3} x - 64 b^{3}$

Step 3.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$16 x = 2 \cdot x \cdot 8$

Step 3.3

Rewrite the polynomial.

$a^{3} (x^{2} - 2 \cdot x \cdot 8 + 8^{2}) - b^{3} x^{2} + 16 b^{3} x - 64 b^{3}$

Step 3.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 8$ .

$a^{3} {(x - 8)}^{2} - b^{3} x^{2} + 16 b^{3} x - 64 b^{3}$

Step 4

Factor $b^{3}$ out of $- b^{3} x^{2} + 16 b^{3} x - 64 b^{3}$ .

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Step 4.1

Factor $b^{3}$ out of $- b^{3} x^{2}$ .

$a^{3} {(x - 8)}^{2} + b^{3} (- 1 x^{2}) + 16 b^{3} x - 64 b^{3}$

Step 4.2

Factor $b^{3}$ out of $16 b^{3} x$ .

$a^{3} {(x - 8)}^{2} + b^{3} (- 1 x^{2}) + b^{3} (16 x) - 64 b^{3}$

Step 4.3

Factor $b^{3}$ out of $- 64 b^{3}$ .

$a^{3} {(x - 8)}^{2} + b^{3} (- 1 x^{2}) + b^{3} (16 x) + b^{3} \cdot - 64$

Step 4.4

Factor $b^{3}$ out of $b^{3} (- 1 x^{2}) + b^{3} (16 x)$ .

$a^{3} {(x - 8)}^{2} + b^{3} (- 1 x^{2} + 16 x) + b^{3} \cdot - 64$

Step 4.5

Factor $b^{3}$ out of $b^{3} (- 1 x^{2} + 16 x) + b^{3} \cdot - 64$ .

$a^{3} {(x - 8)}^{2} + b^{3} (- 1 x^{2} + 16 x - 64)$

Step 5

Factor by grouping.

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Step 5.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot - 64 = 64$ and whose sum is $b = 16$ .

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Step 5.1.1

Factor $16$ out of $16 x$ .

$a^{3} {(x - 8)}^{2} + b^{3} (- 1 x^{2} + 16 (x) - 64)$

Step 5.1.2

Rewrite $16$ as $8$ plus $8$

$a^{3} {(x - 8)}^{2} + b^{3} (- 1 x^{2} + (8 + 8) x - 64)$

Step 5.1.3

Apply the distributive property.

$a^{3} {(x - 8)}^{2} + b^{3} (- 1 x^{2} + 8 x + 8 x - 64)$

Step 5.2

Factor out the greatest common factor from each group.

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Step 5.2.1

Group the first two terms and the last two terms.

$a^{3} {(x - 8)}^{2} + b^{3} ((- 1 x^{2} + 8 x) + 8 x - 64)$

Step 5.2.2

Factor out the greatest common factor (GCF) from each group.

$a^{3} {(x - 8)}^{2} + b^{3} (x (- 1 x + 8) - 8 (- x + 8))$

Step 5.3

Factor the polynomial by factoring out the greatest common factor, $- 1 x + 8$ .

$a^{3} {(x - 8)}^{2} + b^{3} ((- 1 x + 8) (x - 8))$

Step 6

Factor.

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Step 6.1

Rewrite $- 1 x$ as $- x$ .

$a^{3} {(x - 8)}^{2} + b^{3} ((- x + 8) (x - 8))$

Step 6.2

Remove unnecessary parentheses.

$a^{3} {(x - 8)}^{2} + b^{3} (- x + 8) (x - 8)$

Step 7

Combine exponents.

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Step 7.1

Factor $- 1$ out of $- x$ .

$a^{3} {(x - 8)}^{2} + b^{3} (- (x) + 8) (x - 8)$

Step 7.2

Rewrite $8$ as $- 1 (- 8)$ .

$a^{3} {(x - 8)}^{2} + b^{3} (- (x) - 1 (- 8)) (x - 8)$

Step 7.3

Factor $- 1$ out of $- (x) - 1 (- 8)$ .

$a^{3} {(x - 8)}^{2} + b^{3} (- (x - 8)) (x - 8)$

Step 7.4

Rewrite $- (x - 8)$ as $- 1 (x - 8)$ .

$a^{3} {(x - 8)}^{2} + b^{3} (- 1 (x - 8)) (x - 8)$

Step 7.5

Remove parentheses.

$a^{3} {(x - 8)}^{2} + b^{3} \cdot - 1 (x - 8) (x - 8)$

Step 7.6

Raise $x - 8$ to the power of $1$ .

$a^{3} {(x - 8)}^{2} + b^{3} \cdot - 1 ({(x - 8)}^{1} (x - 8))$

Step 7.7

Raise $x - 8$ to the power of $1$ .

$a^{3} {(x - 8)}^{2} + b^{3} \cdot - 1 ({(x - 8)}^{1} {(x - 8)}^{1})$

Step 7.8

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$a^{3} {(x - 8)}^{2} + b^{3} \cdot - 1 {(x - 8)}^{1 + 1}$

Step 7.9

Add $1$ and $1$ .

$a^{3} {(x - 8)}^{2} + b^{3} \cdot - 1 {(x - 8)}^{2}$

Step 8

Factor ${(x - 8)}^{2}$ out of $a^{3} {(x - 8)}^{2} + b^{3} \cdot - 1 {(x - 8)}^{2}$ .

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Step 8.1

Factor ${(x - 8)}^{2}$ out of $a^{3} {(x - 8)}^{2}$ .

${(x - 8)}^{2} a^{3} + b^{3} \cdot - 1 {(x - 8)}^{2}$

Step 8.2

Factor ${(x - 8)}^{2}$ out of $b^{3} \cdot - 1 {(x - 8)}^{2}$ .

${(x - 8)}^{2} a^{3} + {(x - 8)}^{2} (b^{3} \cdot - 1)$

Step 8.3

Factor ${(x - 8)}^{2}$ out of ${(x - 8)}^{2} a^{3} + {(x - 8)}^{2} (b^{3} \cdot - 1)$ .

${(x - 8)}^{2} (a^{3} + b^{3} \cdot - 1)$

Step 9

Rewrite $b^{3} \cdot - 1$ as ${(b \cdot - 1)}^{3}$ .

${(x - 8)}^{2} (a^{3} + {(b \cdot - 1)}^{3})$

Step 10

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

${(x - 8)}^{2} ((a + b \cdot - 1) (a^{2} - a (b \cdot - 1) + {(b \cdot - 1)}^{2}))$

Step 11

Factor.

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Step 11.1

Simplify.

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Step 11.1.1

Move $- 1$ to the left of $b$ .

${(x - 8)}^{2} ((a - 1 b) (a^{2} - a (b \cdot - 1) + {(b \cdot - 1)}^{2}))$

Step 11.1.2

Rewrite $- 1 b$ as $- b$ .

${(x - 8)}^{2} ((a - b) (a^{2} - a (b \cdot - 1) + {(b \cdot - 1)}^{2}))$

Step 11.1.3

Move $- 1$ to the left of $b$ .

${(x - 8)}^{2} ((a - b) (a^{2} - a (- 1 \cdot b) + {(b \cdot - 1)}^{2}))$

Step 11.1.4

Rewrite using the commutative property of multiplication.

${(x - 8)}^{2} ((a - b) (a^{2} - 1 \cdot - 1 a b + {(b \cdot - 1)}^{2}))$

Step 11.1.5

Multiply $- 1$ by $- 1$ .

${(x - 8)}^{2} ((a - b) (a^{2} + 1 a b + {(b \cdot - 1)}^{2}))$

Step 11.1.6

Multiply $a$ by $1$ .

${(x - 8)}^{2} ((a - b) (a^{2} + a b + {(b \cdot - 1)}^{2}))$

Step 11.1.7

Move $- 1$ to the left of $b$ .

${(x - 8)}^{2} ((a - b) (a^{2} + a b + {(- 1 b)}^{2}))$

Step 11.1.8

Rewrite $- 1 b$ as $- b$ .

${(x - 8)}^{2} ((a - b) (a^{2} + a b + {(- b)}^{2}))$

Step 11.1.9

Apply the product rule to $- b$ .

${(x - 8)}^{2} ((a - b) (a^{2} + a b + {(- 1)}^{2} b^{2}))$

Step 11.1.10

Raise $- 1$ to the power of $2$ .

${(x - 8)}^{2} ((a - b) (a^{2} + a b + 1 b^{2}))$

Step 11.1.11

Multiply $b^{2}$ by $1$ .

${(x - 8)}^{2} ((a - b) (a^{2} + a b + b^{2}))$

Step 11.2

Remove unnecessary parentheses.

${(x - 8)}^{2} (a - b) (a^{2} + a b + b^{2})$