## quickstudent one year ago Can someone please check this for me? Factoring Sum of Cubes

1. quickstudent

2. hartnn

3a is incorrect

3. hartnn

which makes 3b and 3c also incorrect

4. quickstudent

Ok, let me review it a moment...

5. hartnn

sure, i was about to say, try to find the error on your own

6. hartnn

and the same mistake is done in 4. you correct one, you correct all!! :)

7. quickstudent

Hmmm, I don't see what's wrong there. Can you explain?

8. hartnn

length = 2t its cube = $$(2t)^3 = 8t^3$$

9. hartnn

basically you forgot the brackets :P

10. quickstudent

Oh, okay. I'll try doing those again then :)

11. hartnn

sure, all 3 a,b,c and 4 a,b,c needs that correction. let me know when you're done, i'll review it again.

12. quickstudent

Here it is now, corrected :) @hartnn

13. hartnn

lets come to 3b now. $$(2t)^3 + 12^3 = (2t+12)((2t)^2 - (2t)\times 12) +12^2$$ makes sense?

14. hartnn

i missed the last bracket $$(2t)^3 + 12^3 = (2t+12)((2t)^2 - (2t)\times 12 +12^2)$$

15. hartnn

got that^^ ?

16. quickstudent

OK, I get it.

17. hartnn

so your answer for 3b should be $$(2t+12)(4t^2 -24t +144)$$

18. hartnn

now for 3c, its (2t)^3 not (8t)^3 ... so (2*7)^3 + 12^3 = 14^3 +12^3 = .........

19. hartnn

same thing goes for 4b and 4c :)

20. hartnn

$$(2t+12)(4t^2 -24t +144)$$ ^^ that can be further factorized $$8(t + 6)(t^2 - 6t + 36)$$

21. quickstudent

OK, now I've correct 3 and 4 again.

22. hartnn

typing error in 4b

23. hartnn

everything else is correct :)

24. quickstudent

I've corrected that typo:) Thanks for helping me:)

25. hartnn

welcome ^_^