Question

Can someone please check this for me?
Factoring Sum of Cubes

Answer 1

3a is incorrect

Answer 2

which makes 3b and 3c also incorrect

Answer 3

Ok, let me review it a moment...

Answer 4

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

Answer 5

and the same mistake is done in 4.

you correct one, you correct all!! :)

Answer 6

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

Answer 7

length = 2t

its cube = \( (2t)^3 = 8t^3 \)

Answer 8

basically you forgot the brackets :P

Answer 9

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

Answer 10

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.

Answer 11

Here it is now, corrected :)

@hartnn

Answer 12

lets come to 3b now.

\((2t)^3 + 12^3 = (2t+12)((2t)^2 - (2t)\times 12) +12^2 \)

makes sense?

Answer 13

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

Answer 14

got that^^ ?

Answer 15

OK, I get it.

Answer 16

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

Answer 17

now for 3c,
its (2t)^3 not (8t)^3 ...

so (2*7)^3 + 12^3 = 14^3 +12^3 = .........

Answer 18

same thing goes for 4b and 4c :)

Answer 19

\((2t+12)(4t^2 -24t +144)\)

^^ that can be further factorized

\(8(t + 6)(t^2 - 6t + 36)\)

Answer 20

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

Answer 21

typing error in 4b

Answer 22

everything else is correct :)

Answer 23

I've corrected that typo:)

Thanks for helping me:)

Answer 24

welcome ^_^