Sum or Difference of Two CubesDate: 10/16/97 at 11:23:38 From: Henderson Subject: Sum or difference of two cubes I am in the tenth grade and I need help with this algebra 2 assignment. I am home schooling and so far I have not had any problems, but I cannot figure out how to come up with the next line for problems like this: Factor and check: c(cubed) + d(cubed) b(cubed) - c(cubed) x(cubed) + 8 y(cubed) - 125 Please show me how to do those and why. The book is going over my head. Thank you very much for your time. - Sarah Date: 10/16/97 at 12:35:03 From: Doctor Bruce Subject: Re: Sum or difference of two cubes Hello Sarah, I'll show you how to factor the first problem: c(cubed) + d(cubed). Then I'll give you a couple of hints about why the other problems are basically the same thing in disguise. See if you can work out the other problems on your own, then. When we have to type math symbols in a message like this, we will usually write c^3 instead of c(cubed), and we use the asterisk (*) to stand for multiplication. Here we go: c^3 + d^3 = (c + d) * (c^2 - c*d + d^2). You should check this! You do that by multiplying out the right side of the equal sign. Several terms will cancel out, and the result will be c^3 + d^3. Go ahead and try it. Now, how is the second problem like the first? This time, we are trying to factor b^3 - c^3. Well, we notice that b^3 - c^3 = b^3 + (-c)^3 Now this fits the pattern of the first problem! Instead of (something)^3 - (something else)^3 we have (something)^3 + (something slightly different)^3 and we already know how to factor that. See if you can "plug" the values of "b" and "-c" into the factorization above. That will give the factorization of b^3 - c^3. The other two problems are special cases of the first two. That's because we can write them x^3 + 8 = x^3 + 2^3 y^3 - 125 = y^3 - 5^3. So, when you factor x^3 + 2^3, "x" takes the place of the "c" in the factorization above, and "2" takes the place of "d". See if you can do them now. -Doctor Bruce, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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