Step 1 :

Rewrite $ a^{6}-64 $ as:

$$ a^{6}-64 = (a^{3})^2 - (8)^2 $$

Now we can apply the difference of squares formula.

$$ I^2 - II^2 = (I - II)(I + II) $$

After putting $ I = a^{3} $ and $ II = 8 $ , we have:

$$ a^{6}-64 = (a^{3})^2 - (8)^2 = ( a^{3}-8 ) ( a^{3}+8 ) $$

Step 2 :

To factor $ a^{3}-8 $ we can use difference of cubes formula:

$$ I^3 - II^3 = (I - II)(I^2 + I \cdot II + II^2) $$

After putting $ I = a $ and $ II = 2 $ , we have:

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

Step 3 :

Step 3: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:

$$ \color{blue}{ b = 2 } ~ \text{ and } ~ \color{red}{ c = 4 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 2 } $ and multiply to $ \color{red}{ 4 } $.

Step 4: Find out pairs of numbers with a product of $\color{red}{ c = 4 }$.

PRODUCT = 4
1     4	-1     -4
2     2	-2     -2

Step 5: Because none of these pairs will give us a sum of $ \color{blue}{ 2 }$, we conclude the polynomial cannot be factored.

Step 4 :

To factor $ a^{3}+8 $ we can use sum of cubes formula:

$$ I^3 - II^3 = (I + II)(I^2 - I \cdot II + II^2) $$

After putting $ I = a $ and $ II = 2 $ , we have:

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

Step 5 :

Step 5: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:

$$ \color{blue}{ b = 2 } ~ \text{ and } ~ \color{red}{ c = 4 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 2 } $ and multiply to $ \color{red}{ 4 } $.

Step 6: Find out pairs of numbers with a product of $\color{red}{ c = 4 }$.

PRODUCT = 4
1     4	-1     -4
2     2	-2     -2

Step 7: Because none of these pairs will give us a sum of $ \color{blue}{ 2 }$, we conclude the polynomial cannot be factored.

Step 6 :

Step 6: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:

$$ \color{blue}{ b = -2 } ~ \text{ and } ~ \color{red}{ c = 4 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -2 } $ and multiply to $ \color{red}{ 4 } $.

Step 7: Find out pairs of numbers with a product of $\color{red}{ c = 4 }$.

PRODUCT = 4
1     4	-1     -4
2     2	-2     -2

Step 8: Because none of these pairs will give us a sum of $ \color{blue}{ -2 }$, we conclude the polynomial cannot be factored.

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