Question

heya!
solve this please:
#explanatory answers needed
if a=x-y,
b=y-z,
c=z-x
then (x-y)³+(y-z)³+(z-x)³/3 (x-y)(y-z)(z-x)=?
#ntse-preparation

Answer 1

These are the tricky questions,which makes you to think as harder ones and you will feel it time taking. But these are easy. A brilliant student takes 20-30 seconds.

Now, Going back to the question.

a=x-y,
b=y-z,
c=z-x

Here, a+b+c
=>x-y+y-z+z-x
=> 0.

We know that if a+b+c =0 then a³+b³+c³=3abc

[ a+b+c =0
=> a+b =-c
cubing on both sides,
(a+b) ^3 =-c^3
a³+b³+3ab(a+b)=-c³
wkt a+b=-c
a³+b³+3ab(-c)=-c³
a³+b³-3abc=-c³
a³+b³+c³=3abc ]

This is just for a reference, steps inside [ ] aren't required to be learnt.

Now back to question.

So here a+b+c =0

a³+b³+c³=3abc

replacing a,b,c by their values.

(x-y)³+(y-z)³+(z-x)³= 3 (x-y)(y-z)(z-x)

transposing the terms

we get (x-y)³+(y-z)³+(z-x)³/3 (x-y)(y-z)(z-x)=1

Answer 2

Step-by-step explanation:

hope it may be helpful..