RRB Junior Engineer Stage 1 Model Papers 6

© examsiri.com

Question : 24 of 99

Marks: +1, -0

If

(2a-1)^2+(4b-3)^2+(4c+5)^2=0

then the value of

{a^3+b^3+c^3-3abc}/{a^2+b^2+c^2}

is

$1 3/8$
$3 3/8$
$2 3/8$
0

Solution:

(2a-1)^2+(4b-3)^2+(4c+5)^2=0

=2a-1=0;a=1/2;4b-3=0;b=3/4;4c+5=0c={-5}/4a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)

But

a + b + c = 0

So,

a^3+b^3+c^3-3abc=0

So,

{a^3+b^3+c^3-3abc}/{a^3+b^3+c^3}=0

© examsiri.com

Go to Question:

Prev Question

Next Question