Question

Question below:

Answer 1

@whpalmer4

Answer 2

@AccessDenied , @ganeshie8

Answer 3

@mathmale Can you try this?

Answer 4

well did you try replacing \[a \text{ with } A_1+B_1i \text{ and } b \text{ with } A_2+B_2i \text{ and } c \text{ with } A_3+B_3i ,
\\ A_1,A_2,A_3,B_1,B_2,B_3 \\ \text{ are real numbers}\]

I don't know if this helps but we can try to see.

Answer 5

ir I wonder what would happen if we just square both sides of that one equation

Answer 6

I would try that last thing first

Answer 7

Substituting A1+B1i seems to be a long process. I remember doing a similar problem before. In that the property used was z (z bar)=(mod z)^2. But I dont remember how they used it.

Answer 8

\[x+y+z=0 => (x+y+z)^2=0^2 => \\ ((x+y)+z)^2=(x+y)^2+2z(x+y)+z^2 \\ =x^2+2xy+y^2+2zx+2zy+z^2 \\= x^2+y^2+z^2+2xy+2zx+2zy \\=x^2+y^2+z^2+2(xy+zx+zy)=0 \]
=>
\[x^2+y^2+z^2=-2(xy+zx+zy)\]

Answer 9

Then this is the part where I guess more thinking is needed.

Answer 10

combine fractions on the right hand side

Answer 11

multiply everything out on top and bottom
I think we should see something

Answer 12

yep

Answer 13

that wasn't too bad
did you get it yet?

Answer 14

if you do what I said in latex and then do the whole combining of fractions on the right hand side
and then do all the multiplying on top and bottom
you will see a nice surprise

Answer 15

Sorry I was away. I will take a look at that now.

Answer 16

Do you see what I'm talking about yet?

Answer 17

Doing simplification steps.

Answer 18

I am stuck here:

\[R.E.=\dfrac{-2b}{c-a} \times( \dfrac{(a^2-ab+bc-c^2)}{(b-c)(a-b)}+\dfrac{ca}{(a-b)(b-c)})\]

Answer 19

well what did you do..

Answer 20

i mean whoa

Answer 21

Substituted for xy,yz,zx and simplified.

Answer 22

Oh wait I think I made a terrible mistake :P

Answer 23

\[x+y+z=0 => (x+y+z)^2=0^2 => \\ ((x+y)+z)^2=(x+y)^2+2z(x+y)+z^2 \\ =x^2+2xy+y^2+2zx+2zy+z^2 \\= x^2+y^2+z^2+2xy+2zx+2zy \\=x^2+y^2+z^2+2(xy+zx+zy)=0 \]

\[x^2+y^2+z^2=-2(xy+zx+zy) \]


So we have
\[-2(\frac{a}{b-c}\frac{b}{c-a}+\frac{a}{b-c}\frac{c}{a-b}+\frac{b}{c-a}\frac{c}{a-b})\]

You need to find a common denominator
then write the fractions as one

Answer 24

\[-2(\frac{a}{b-c}\frac{b}{c-a} \frac{a-b}{a-b}+\frac{a}{b-c}\frac{c}{a-b}\frac{c-a}{c-a}+\frac{b}{c-a}\frac{c}{a-b}\frac{b-c}{b-c})\]
Now write the fractions as one

Answer 25

multiply everything on top out
multiply everything on bottom out

Answer 26

Should I again regroup on numerator?

Answer 27

I want you to multiply everything out on top
a*b*(a-b)
also
a*c*(c-a)
also
b*c*(b-c)

Answer 28

Which gives me:

\[\dfrac{a^2b-b^2a+b^2c-bc^2+c^2+c^2a-ca^2}{(c-a)(b-c)(a-b)}\]

Answer 29

*neglect that c^2 typo

Answer 30

now multiply the bottom

Answer 31

a^2b +b^2c -ab^2 +ac^2 -a^2c -bc^2

Answer 32

Now what to do? :O

Answer 33

ok look at the top and look at the bottom
what do you notice?

Answer 34

Identical!!

Answer 35

Also did you distribute that 2 a long time ago or not?

Answer 36

i mean that negative?

Answer 37

No so -2 is answer.

Answer 38

I think something happen to a negative?
Because I got 2.

Answer 39

I will check my answer again
You check yours too

Answer 40

I think your answer is right since options are all positive.

Answer 41

maybe you didn't multiply the bottom correctly

Answer 42

or maybe everything looked so similar you thought it was identitcal

Answer 43

and it was the opposite

Answer 44

Oh okay.

Answer 45

Thanks for the help!!

Answer 46

\[\frac{ab(a-b)+ac(c-a)+bc(b-c)}{(b-c)(c-a)(a-b)} \\ \frac{a^2b-ab^2+ac^2-a^2c+b^2c-bc^2}{(bc-ab-c^2+ca)(a-b)} \\ \frac{a^2b-ab^2+ac^2-a^2c+b^2c-bc^2}{(abc-b^2c-a^2b+ab^2-c^2a+c^2b+ca^2-acb)} \\ \frac{a^2b-ab^2+ac^2-a^2c+b^2c-bc^2}{-a^2b+ab^2-ac^2+a^2c-b^2c+bc^2}\]

Answer 47

so we have -2(-1)=2

Answer 48

Okay :)

Answer 49

I never expected the denominator to come out like that!

Answer 50

let me try once. hihihi
\[\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\]
\[(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})^2=0\]
\[((\frac{a}{b-c}+\frac{b}{c-a})+\frac{c}{a-b})^2=0\]
open the bracket
\[(\frac{a}{b-c}+\frac{b}{c-a})^2+2(\frac{a}{b-c}+\frac{b}{c-a})*\frac{c}{a-b}+(\frac{c}{a-b})^2=0\]
\[(\frac{a}{b-c})^2+2\frac{ab}{(b-c)(c-a)}+(\frac{b}{c-a})^2+2(\frac{a}{b-c}+\frac{b}{c-a})*\frac{c}{a-b}+(\frac{c}{a-b})^2=0\]

so that , \[(\frac{a}{b-c})^2+(\frac{b}{c-a})^2+(\frac{c}{a-b})^2=-2\frac{ab}{(b-c)(c-a)}-2(\frac{a}{b-c}+\frac{b}{c-a}) \]
\[(\frac{a}{b-c})^2+(\frac{b}{c-a})^2+(\frac{c}{a-b})^2 =-2\frac{ab}{(b-c)(c-a)}-2(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b} )-2\frac{c}{a-b} \]

\[(\frac{a}{b-c})^2+(\frac{b}{c-a})^2+(\frac{c}{a-b})^2 =-2\frac{ab}{(b-c)(c-a)}-2\frac{c}{a-b} \]

Answer 51

I got 2 as well, but I'm not going to show my work because I had Mathematica do it for me :-)