SSC CHSL 15 Nov 2015 Shift 2 Paper

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Question : 115 of 200

Marks: +1, -0

If x=a(b-c), y=b(c-a), z=c(a-b), then the value of

(x /a )^3 + (y/ b ) ^3+ ( z/c ) ^3

is :

${xyz}/{abc}$
0
${3xyz}/{abc}$
${2xyz}/{abc}$

Solution:

Given :

x = a(b − c) y = b(c − a) z = c(a − b)

⇒

x/a=(b − c)

-----(i)
and

y/b=(c− a)

------(ii)
and

z/c=(a-b)

----(iii)
Adding equations (i), (ii) and (iii), we get :
⇒

x/a + y/b + z/c

= (b − c) + (c − a) + (a − b)

⇒

x/a + y/b + z/c=0

Now, we know that if

(p+q+r)=0

then

p^3+q^3+r^3=3pqr

∴

(x/a) ^3+ ( y/b)^3 + (z/c)^3

=

3×(x/a)× (y/b)× (z/c)

=

{3xyz}/{abc}

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