# Math Help - a^4+b^4+c^4

1. ## a^4+b^4+c^4

Suppose a,b,c satisfy the following equation:
a+b+c=3;
a2+ b2 +c2= 5;
a3+b3+c3=7;
what is a4+b4+c4

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2. Originally Posted by kamaksh_ice
Suppose a,b,c satisfy the following equation:
a+b+c=3;
a2+ b2 +c2= 5;
a3+b3+c3=7;
what is a4+b4+c4

SEND ME DE ENTIRE SOLUTION
Hello,

I assume that you mean a² when you write a2.

I can't give you the complete solution, only some considerations. But maybe you can use them for a start:

$(a+1)^4+(b+1)^4+(c+1)^4$ = $a^4+4a^3+6a^2+4a+1 + b^4+4b^3+6b^2+4b+1+c^4+4c^3+6c^2+4c+1$

$a^4+b^4+c^4+4(a^3+b^3+c^3)+6(a^2+b^2+c^2)+4(a+b+c) +3$ According to the problem this term can be reduced to:

$a^4+b^4+c^4+4 \cdot (7)+6 \cdot (5)+4 \cdot (3)+3 = a^4+b^4+c^4+73$

And now I need a small spark of inspiration.

3. $a^2+b^2+c^2 = 5$
Square,
$a^4+b^4+c^4 + 2(a^2b^2+b^2c^2+a^2c^2)=25$...(1)

$a+b+c=3$
Square,
$a^2+b^2+c^2+2(ab+ac+bc)=9$
Thus,
$ab+bc+ac=2$
Square,
$a^2b^2+b^2c^2+a^2c^2+2(a^2bc+ab^2c+abc^2)=4$
Thus,
$a^2b^2+b^2c^2+a^2c^2+2abc(a+b+c)=4$
Thus,
$a^2b^2+b^2c^2+a^2c^2+6abc = 4$...(2)

Use the identity,
$a^3+b^3+c^3 - 3abc = (a^2+b^2+c^2 - ab - bc - ac)(a+b+c)$
Thus,
$7 - 3abc = (5-2)(3)$
Thus,
$6abc=-4$...(3)

Thus by (2) and (3) we have,
$a^2b^2+b^2c^2+a^2c^2=8$

Thus by (1) we have,
$a^4+b^4+c^4+2(8)=25$
Thus,
$a^4+b^4+c^4=9$

4. Originally Posted by kamaksh_ice
Suppose a,b,c satisfy the following equation:
a+b+c=3;
a2+ b2 +c2= 5;
a3+b3+c3=7;
what is a4+b4+c4
You may, of course, solve the system by substitution. Unfortunately when I do this I can't get an exact answer for a without using Cardano's method.

-Dan