not so easy

Algebra Level 1

if $a+b+c=0$ then evaluate

$a^4/(b+c)+b^4/(c+a)+c^4/(a+b)+3abc$ .

The answer is 0.

4 solutions

Akash Deep
Aug 2, 2014

i am just wondering why was the name of this problem not so easy

Exactly, could be done mentally also!:D

Anik Mandal - 6 years, 10 months ago

Yeah! Same here!Just to confuse, I guess!

Kartik Sharma - 6 years, 10 months ago

i think this is from a set called "not so easy" because i noticed a problem with same title. @akash @kartik and @anik

Mardokay Mosazghi - 6 years, 10 months ago

Rishi Hazra
Aug 1, 2014

replacing the denominators

b+c=-a

c+a=-b

a+b=-c

we get -(a^3 +b^3 +c^3) +3abc-----------------eqn(1)

and

(a+b+c)^3=0=(a+b)^3 +c^3+3 (a+b) c*0

=a^3 +b^3 +c^3 +3ab(a+b)

writng (a+b)=-c

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

thus the ans to eqn(1)=0

how to find the remainder when 128^1000 is divided by 153 ??.

Santosh Tiwari - 6 years, 10 months ago

it can be done more easily whenever a condition of a+b+c = 0 arises then in that case a^3 + b^3 +c^3 = 3abc then when the equation was like -(a^3+b^3+c^3)+3abc it could be written as -3abc +3abc which is equal to 0

Palash Som - 6 years, 10 months ago

Aswad Hariri Mangalaeng
Sep 1, 2014

a+b+c=0 So, a+b=-c, a+c=-b, and b+c=-b.. Subtituted at question,,,

Praveen Tiwari
Aug 9, 2014

not a tough problem..............

not so easy

The answer is 0.

4 solutions

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