## ParthKohli 2 years ago Factor.

1. ParthKohli

2. ParthKohli

I know this simplifies to$\left(\dfrac{2}{3}x + \dfrac{5}{3}y\right)^3 + \rm blah$

3. ParthKohli

So for the first two terms, should I use $$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$?

4. ParthKohli

That'd make stuff very long.

5. hartnn

its like $$(a+b)^3+(c-b)^3-(a+c)^3$$

6. ParthKohli

Yeah. :-|

7. ParthKohli

Wait, so$(a + c)(a^2 + 2ab + b^2 - (a+b)(c -b) + c^2 - 2cb + b^2) - (a + c)^3$

8. hartnn

$$\large = 3(a+b)(a+c)(b-c)$$

9. ParthKohli

How did you get that?

10. ParthKohli

It's correct, but is it some sort of identity taught?

11. hartnn

ok, first take last 2 terms, (c-b)^3-(a+c)^3 = ... ? use A^3-B^3 =...

12. hartnn

not identity, you need to work that out...

13. hartnn

its quite long though

14. ParthKohli

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$So here,$= -(b +a)(\cdots)$Err

15. ParthKohli

No shortcuts?

16. hartnn

none that i can think of.

17. ParthKohli

There's always a shortcut in exam questions. It's not like an Olympiad one.

18. hartnn

then you must be having choices ?

19. ParthKohli

No choices given. That's all.

20. ParthKohli

:-(

21. hartnn

then i'll wait for someone to suggest a shortcut.....

22. ParthKohli

OK, thanks for the support :-)

23. some_someone

Is Pascal's Triangle any useful? http://www.mathsisfun.com/pascals-triangle.html

24. some_someone

I usually use that, but i don't know if its going to help you?

25. ParthKohli

@hartnn Oh BTW, I found the solution.

26. hartnn

any shortcuts ?

27. ParthKohli
28. hartnn

isn't that exactly what i told you ?

29. ParthKohli

:-P