a's, b's and c's

Algebra Level 3

If a + b + c = 0 a+b+c = 0 and a b c ≠ 0 abc \neq 0 , then find the value of a 4 + b 4 + c 4 a 2 b 2 + b 2 c 2 + c 2 a 2 \frac { { a }^{ 4 } + { b }^{ 4 } + { c }^{ 4 } } { { a }^{ 2 }{ b }^{ 2 }+ { b }^{ 2 }{ c }^{ 2 }+ { c }^{ 2 }{ a }^{ 2 } } .


The answer is 2.

Tanveen Dhingra by tanveen dhingra , 19, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Yang Cheng
Oct 7, 2014

a 4 + b 4 + c 4 a 2 b 2 + b 2 c 2 + c 2 a 2 = ( a 2 + b 2 + c 2 ) 2 − 2 ( a 2 b 2 + b 2 c 2 + c 2 a 2 ) a 2 b 2 + b 2 c 2 + c 2 a 2 = [ ( a + b + c ) 2 − 2 ( a b + b c + c a ) ] 2 − 2 ( a 2 b 2 + b 2 c 2 + c 2 a 2 ) a 2 b 2 + b 2 c 2 + c 2 a 2 = 4 ( a b + b c + c a ) 2 − 2 ( a 2 b 2 + b 2 c 2 + c 2 a 2 ) a 2 b 2 + b 2 c 2 + c 2 a 2 ∵ a + b + c = 0 = 4 [ a 2 b 2 + b 2 c 2 + c 2 a 2 + 2 a b c ( a + b + c ) ] − 2 ( a 2 b 2 + b 2 c 2 + c 2 a 2 ) a 2 b 2 + b 2 c 2 + c 2 a 2 = 2 \dfrac{a^4+b^4+c^4}{a^2b^2+b^2c^2+c^2a^2} \\ = \dfrac{(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)}{a^2b^2+b^2c^2+c^2a^2}\\ =\dfrac{[(a+b+c)^2-2(ab+bc+ca)]^2-2(a^2b^2+b^2c^2+c^2a^2)}{a^2b^2+b^2c^2+c^2a^2}\\ =\dfrac{4(ab+bc+ca)^2-2(a^2b^2+b^2c^2+c^2a^2)}{a^2b^2+b^2c^2+c^2a^2} \qquad ∵a+b+c=0 \\\\ =\dfrac{4[a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)]-2(a^2b^2+b^2c^2+c^2a^2)}{a^2b^2+b^2c^2+c^2a^2}\\ =2

8 Helpful 0 Interesting 0 Brilliant 0 Confused

The answer can also be 0.

Anuj Shikarkhane - 6 years, 7 months ago

Log in to reply

If the answer is 0 it means that a=b=c =0 If this so then by putting the values the denominator becomes 0 and division by 0 is not possible.

tanveen dhingra - 6 years, 7 months ago

As Tanveen points out, the expression cannot be equal to 0. In the worst case scenario, it is undefined.

I have edited the problem slightly for clarity.

Calvin Lin Staff - 6 years, 6 months ago
Mehul Chaturvedi
Dec 11, 2014

We can Let a = 1 a=1 b = 1 b=1 and c = − 2 c=-2

or a = 1 a=1 b = 2 b=2 and c = − 3 c=-3

Means any a , b , c a,b,c which on adding results to 0 0

3 Helpful 0 Interesting 0 Brilliant 0 Confused

during competitive exams this is the best way to do it [make sure your assumptions match the conditions given ]

Abu Zubair - 6 years, 5 months ago
Jared Low
Jan 2, 2015

Rearrange the first equation to get a = − ( b + c ) a=-(b+c) and substitue this into our given expression to get:

( − ( b + c ) ) 4 + b 4 + c 4 b 2 c 2 + ( b 2 + c 2 ) ( ( − ( b + c ) ) 2 ) \Large{\frac{(-(b+c))^4+b^4+c^4}{b^2c^2+(b^2+c^2)((-(b+c))^2)}}

= 2 b 4 + 4 b 3 c + 6 b 2 c 2 + 4 b c 3 + 2 c 4 b 2 c 2 + b 4 + 2 b 3 c + 2 b 2 c 2 + 2 b c 3 + c 4 \Large{=\frac{2b^4+4b^3c+6b^2c^2+4bc^3+2c^4}{b^2c^2+b^4+2b^3c+2b^2c^2+2bc^3+c^4}}

= 2 ( b 4 + 2 b 3 c + 3 b 2 c 2 + 2 b c 3 + c 4 ) b 4 + 2 b 3 c + 3 b 2 c 2 + 2 b c 3 + c 4 = 2 \Large{=\frac{2(b^4+2b^3c+3b^2c^2+2bc^3+c^4)}{b^4+2b^3c+3b^2c^2+2bc^3+c^4}=\boxed{2}}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...