Question 156973
Do you want to factor this?



Looking at {{{8x^6-7x^3y^3-y^6}}} we can see that the first term is {{{8x^6}}} and the last term is {{{-y^6}}} where the coefficients are 8 and -1 respectively.


Now multiply the first coefficient 8 and the last coefficient -1 to get -8. Now what two numbers multiply to -8 and add to the  middle coefficient -7? Let's list all of the factors of -8:




Factors of -8:

1,2,4,8


-1,-2,-4,-8 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to -8

(1)*(-8)

(2)*(-4)

(-1)*(8)

(-2)*(4)


note: remember, the product of a negative and a positive number is a negative number



Now which of these pairs add to -7? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -7


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">-8</td><td>1+(-8)=-7</td></tr><tr><td align="center">2</td><td align="center">-4</td><td>2+(-4)=-2</td></tr><tr><td align="center">-1</td><td align="center">8</td><td>-1+8=7</td></tr><tr><td align="center">-2</td><td align="center">4</td><td>-2+4=2</td></tr></table>



From this list we can see that 1 and -8 add up to -7 and multiply to -8



Now looking at the expression {{{8x^6-7x^3y^3-y^6}}}, replace {{{-7x^3y^3}}} with {{{x^3y^3+-8x^3y^3}}} (notice {{{x^3y^3-8x^3y^3}}} adds up to {{{-7x^3y^3}}}. So it is equivalent to {{{-7x^3y^3}}})


{{{8x^6+highlight(x^3y^3-8x^3y^3)-y^6}}}



Now let's factor {{{8x^6+x^3y^3-8x^3y^3-y^6}}} by grouping:



{{{(8x^6+x^3y^3)+(-8x^3y^3-y^6)}}} Group like terms



{{{x^3(8x^3+y^3)-y^3(8x^3+y^3)}}} Factor out the GCF of {{{x^3}}} out of the first group. Factor out the GCF of {{{-y^3}}} out of the second group



{{{(x^3-y^3)(8x^3+y^3)}}} Since we have a common term of {{{8x^3+y^3}}}, we can combine like terms



So {{{8x^6-7x^3y^3-y^6}}} factors to {{{(x^3-y^3)(8x^3+y^3)}}}



{{{(x-y)(x^2+xy+y^2)(2x+y)(4x^2-2xy+y^2)}}} Now factor {{{x^3-y^3}}} to get {{{(x-y)(x^2+xy+y^2)}}} (use the difference of cubes formula) and factor {{{8x^3+y^3}}} to get {{{(2x+y)(4x^2-2xy+y^2)}}} (use the sum of cubes formula)



{{{(x-y)(2x+y)(x^2+xy+y^2)(4x^2-2xy+y^2)}}} Rearrange the factors



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     Answer:

So {{{8x^6-7x^3y^3-y^6}}} completely factors to {{{(x-y)(2x+y)(x^2+xy+y^2)(4x^2-2xy+y^2)}}}