Factoring the Difference of Cubes

u12_l2_t3_we2 Factoring the Difference of Cubes

[kerriann]

You are so talented at what you do! Thank you for making my life soo much easier!!

[hrperformance]

i think it should be +2cd instead of -2cd in the final form

[aliasghar]

we believe you, thanks for all your efforts..

[Kradle]

I think I found an error...this is a difference of cubes; however the formula he proceeds with is for the Sum of cubes (a-b)(a^2+ab+b^2) --- NOT (a-b)(a^2-ab+b^2), which is what he used...so shouldn't it be 5(2c-d)(4c^2+2cd+d^2)?

[fshinsa]

Wouldn't it be 5(2c - d)(4c² + 2cd + d²) { a "+" +2cd² not -2cd² ] }

Looking at the a³ - b³ = (a -b)(a² + ab + b²) .

From what I figured, from your solution, would be 40c³ - 40c²d + 20cd² - 5d³, not back to the orginal problem of 40c³ - 5d³ .

Just something I noticed.

[hrperformance]

I don't understand how you get from a3-b3 to the difference of cubes identity! how did the first person do that???
I totally except it is correct (which you can check by expanding the identity and then cancelling terms), but how is it done step by step?

[aliasghar]

SOMEONE POINTED OUT: I think I found an error...this is a difference of cubes; however the formula he proceeds with is for the Sum of cubes (a-b)(a^2+ab+b^2) --- NOT (a-b)(a^2-ab+b^2), which is what he used...so shouldn't it be 5(2c-d)(4c^2+2cd+d^2)?

[alexisroque]

does someone know of a good online Tutor? please i need this class to graduate and i don't understand this class