IF: $\large{\frac{x}{a}=\frac{y}{b}+\frac{z}{c}}$ Then show that :
$$\large{\frac{x^3+a^3}{x^2+a^2}+\frac{y^3+b^3}{y^2+b^2}+\frac{z^3+c^3}{z^2+c^2}=\frac{(x+y+z)^3+(a+b+c)^3}{(x+y+z)^2+(a+b+c)^2}}$$

Question

IF: $\large{\frac{x}{a}=\frac{y}{b}+\frac{z}{c}}$ Then show that : 
$$\large{\frac{x^3+a^3}{x^2+a^2}+\frac{y^3+b^3}{y^2+b^2}+\frac{z^3+c^3}{z^2+c^2}=\frac{(x+y+z)^3+(a+b+c)^3}{(x+y+z)^2+(a+b+c)^2}}$$

anonymous · Answer

Nice question..

Can't this be more difficult ??

mathslover · Answer

x/a = y/b = z/c = k 

I tried like this : @waterineyes :D

anonymous · Answer

There is plus sign in between or they all are equal ??

mathslover · Answer

there is plus

anonymous · Answer

Can you put then equal then ???

anonymous · Answer

*them

mathslover · Answer

nope

mathslover · Answer

How we will put equal  @waterineyes ?:

well .. this is what i did

anonymous · Answer

I am thinking what we can do in it..

mathslover · Answer

x=ak
y=bk
z=ck ?

mathslover · Answer

@vishweshshrimali5 @TuringTest @asnaseer @KingGeorge

anonymous · Answer

You cannot put equal they are in plus form..

mathslover · Answer

no no .. :
x/a = y/b = z/c = k

x=ak , y = bk and z = ck

anonymous · Answer

k = k + k        No they are not equal..

mathslover · Answer

what? k = k + k ???

anonymous · Answer

You are letting them as k..

mathslover · Answer

yea :

let x/a = y/b = z/c = k

anonymous · Answer

See I show you..

mathslover · Answer

k

anonymous · Answer

$$\large \frac{x}{a}=\frac{y}{b}+\frac{z}{c} \implies k \ne k + k     \;\;  ( Agree?)$$

mathslover · Answer

yes it implies that : 

k is not equal to 2k

anonymous · Answer

So how can you put them as k ??

anonymous · Answer

x/a is sum of y/b and z/c..

mathslover · Answer

Oh wait .. :from your side a slap to me :( 
mistake there:

x/a = y/b = z/c is the correct situation of "if"

anonymous · Answer

If they were equal then you can put them as k..

mathslover · Answer

yes but wht next

anonymous · Answer

I am still thinking..

mathslover · Answer

ok

anonymous · Answer

We will need here more mathematicians..

@mukushla can you help here..

mathslover · Answer

right have to go now.. sorry 

and please think abt this more

anonymous · Answer

Yeah don't worry..

I am working on it..

anonymous · Answer

oh man thats equal not plus...i think we did it before...

anonymous · Answer

Really ??

anonymous · Answer

mathslover is offline..

We have to solve it somewhat..

anonymous · Answer

Yes they are equal..

anonymous · Answer

http://openstudy.com/study#/updates/501aa609e4b02742c0b1f703

anonymous · Answer

@mathslover you have solved half part Now solve for RHS.


In RHS just replace x = ak, y = bk and z = ck

After factoring you will get : $(a + b + c) \times \frac{k^3 + 1}{k^2 + 1}$ which is equal to LHS..

anonymous · Answer

yeah... and now we just need to put$$k=\frac{x+y+z}{a+b+c}$$

anonymous · Answer

But if we will solve Somewhat for LHS and somewhat for RHS then also we can prove them equal if they are coming equal too..


@mathslover your question should be actually like this:
If 
$$\frac{x}{a} = \frac{y}{b} = \frac{c}{z}$$

Then .......................

mathslover · Answer

that is what i am thinking then?

anonymous · Answer

I think you posted the same question..

mathslover · Answer

Yes but didn't get the answer

anonymous · Answer

How??

anonymous · Answer

You have done half part correctly..

anonymous · Answer

Now change the value of k according to mukushla gave above..

mathslover · Answer

OH k =(x+y+z)/(a+b+c) 

but i had taken k = x/a = y/b = z/c 

so how it will be k = (x+y+z)/(a+b+c)

anonymous · Answer

$$k = \frac{x}{a}=\frac{y}{b}=\frac{z}{c} \;\; then \;\; k = \frac{x + y + z}{a + b + c}$$

anonymous · Answer

For example:

$$k= \frac{1}{2} = \frac{2}{4} = \frac{3}{6} \implies k = \frac{1 + 2 + 3}{2 + 4 + 6} \implies k = \frac{6}{12} = \frac{1}{2}$$

mathslover · Answer

right thanks a lot i got it .. 

thanks @mukushla and @waterineyes

anonymous · Answer

Welcome dear..

anonymous · Answer

let me complete the solution$$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k$$ so$$x=ak$$$$y=bk$$$$z=ck$$add them
$$x+y+z=k(a+b+c)$$so$$k=\frac{x+y+z}{a+b+c}$$

anonymous · Answer

That makes sense in a better and understandable way..

mathslover · Answer

thanks @mukushla that helped

anonymous · Answer

np...man

jiteshmeghwal9 · Answer

similar questions i also gt in my N.T.S.E. exam bt i gt confused whereas they were too easy to solve , HAHAHA :)

mathslover · Answer

that is the speciality of NTSe

jiteshmeghwal9 · Answer

yes!

anonymous · Answer

$$
x=a k\
y=b k\
z= c k \
\frac{a^3+x^3}{a^2+x^2}+\frac{b^3+y^3}{b^2+y^2}+\frac{c^3+z^3}{c^2+z^2}=\
\frac{a^3 k^3+a^3}{a^2
   k^2+a^2}+\frac{b^3 k^3+b^3}{b^2
   k^2+b^2}+\frac{c^3 k^3+c^3}{c^2
   k^2+c^2}=\
\frac{\left(k^3+1\right)
   (a+b+c)}{k^2+1}\

$$
The other side
$$
\frac{(a+b+c)^3+(x+y+z)^3}{(a+b+c)^2+(x+y+z)^2}=\
\frac{(a k+b k+c k)^3+(a+b+c)^3}{(a k+b k+c
   k)^2+(a+b+c)^2}=\
\frac{\left(k^3+1\right)
   (a+b+c)}{k^2+1}


$$