identity qn (1 Viewer)

[bos1234]

is there a short cut for this qn instead of expanding?

-------------------------------------------------------------------------------------------

here is a bit of a tough one :S

 

[falcon07]

The first one:
a=2, as ax^2 is the only term that will give you terms to the 4th power.
If x = 0, c^2 = 25, therefore c = +-5
If x = 1, 4 - 12 - 11 + 30 + 25 = 36 = (2 + b +- 5)^2
b = -1, 9, -9 or -3? (I'm not sure if there should be that many solutions, but the general method is to sub in values for x that simplify the problem, such as x=1 and x=0)

2nd: Let x = -a and half the terms on the left side cancel out.

3rd: Multiplying by abc(a+1)(b+1)(c+1) gives you the 2nd identity, but with x=1. You have already proven that the 2nd identity is true for any value of x.

 

[bos1234]

i get the 1st qn..

i dont understand the 2nd questions though...
is this genrally what i do??

sub in -a and cancel. Then what can i do?? How can i show its identical to x??

thanks for the working out
bye

 

[Riviet]

2) Remember that substituting x=-a into RHS gives -a. On the LHS, you can get a common denominator by taking out a negative sign out of one of the denominators ie -(b-c) or -(c-b). Factorise the numerator which allows you to further cancel out to give -a.

edit: please see jyu's post

 

[jyu]

The first one:
a=2, as ax^2 is the only term that will give you terms to the 4th power.
If x = 0, c^2 = 25, therefore c = +-5
If x = 1, 4 - 12 - 11 + 30 + 25 = 36 = (2 + b +- 5)^2
b = -1, 9, -9 or -3? (I'm not sure if there should be that many solutions, but the general method is to sub in values for x that simplify the problem, such as x=1 and x=0)

2nd: Let x = -a and half the terms on the left side cancel out.

3rd: Multiplying by abc(a+1)(b+1)(c+1) gives you the 2nd identity, but with x=1. You have already proven that the 2nd identity is true for any value of x.

First
a = 2, b = -3 and c = -5

OR a = -2, b = 3 and c = 5

:wave:

 

[bos1234]

fanx i got it!!!!!!!!!!!!

 

[jyu]

2) Remember that substituting x=-a into RHS gives -a. On the LHS, you can get a common denominator by taking out a negative sign out of one of the denominators ie -(b-c) or -(c-b). Factorise the numerator which allows you to further cancel out to give -a.

So it is true for x = -a.
Is it true for all real x?

Use algebraic manipulation to simplify the left side to x.

bc(a+x)/[(a-b)(a-c)] + ......

= -bc(a+x)(b-c)/[(a-b)(b-c)(c-a)] + ........ etc

:wave: