In Exercises 75–94, factor using the formula for the sum or diffe...

Question

In Exercises 75–94, factor using the formula for the sum or difference of two cubes. (x − y)³ − y³

Accepted Answer

Characterize the expression completely R minus two S to the third minus S to the third. Now, with this expression, we actually notice this is a difference of perfect cubes of the form A to the third minus B to the third where A Is RR -2 SB is just S Now the ident in these states, there are difference of two Cubes can be rewritten as two factors A minus B multiplied by A squared plus A B plus B squared. Now we can go ahead and substitute our A and B into this identity B K R minus two S minus S four A minus B for a second factor R -2 S Squared Plus R -2 S multiplied by S plus as quick. The first factor simplifies the R -3 s. Now the second factor we have R minus two S squared which simplifies to A -4 RS Plus four S Squared. A middle term here simplifies RS -2 S Squared. And finally add on S squared at the end R -3 S is then multiplied by R squared -3 RS Plus three S quick, we can't factor this out anymore. So our final answer R minus three S multiplied by R squared minus three R S plus three S squared corresponds with answer A OK. I hope to help you solve the problem. Thanks for watching. Goodbye.

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In Exercises 75–94, factor using the formula for the sum or difference of two cubes. (x − y)³ − y³

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