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

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- Jul 26th 2007, 04:55 AM #1

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- Jul 26th 2007, 05:59 AM #2
Hello,

I assume that you mean aČ when you write a2.

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

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

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

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

And now I need a small spark of inspiration.

- Jul 26th 2007, 07:55 AM #3

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$\displaystyle a^2+b^2+c^2 = 5$

Square,

$\displaystyle a^4+b^4+c^4 + 2(a^2b^2+b^2c^2+a^2c^2)=25$...(1)

$\displaystyle a+b+c=3$

Square,

$\displaystyle a^2+b^2+c^2+2(ab+ac+bc)=9$

Thus,

$\displaystyle ab+bc+ac=2$

Square,

$\displaystyle a^2b^2+b^2c^2+a^2c^2+2(a^2bc+ab^2c+abc^2)=4$

Thus,

$\displaystyle a^2b^2+b^2c^2+a^2c^2+2abc(a+b+c)=4$

Thus,

$\displaystyle a^2b^2+b^2c^2+a^2c^2+6abc = 4$...(2)

Use the identity,

$\displaystyle a^3+b^3+c^3 - 3abc = (a^2+b^2+c^2 - ab - bc - ac)(a+b+c)$

Thus,

$\displaystyle 7 - 3abc = (5-2)(3)$

Thus,

$\displaystyle 6abc=-4$...(3)

Thus by (2) and (3) we have,

$\displaystyle a^2b^2+b^2c^2+a^2c^2=8$

Thus by (1) we have,

$\displaystyle a^4+b^4+c^4+2(8)=25$

Thus,

$\displaystyle a^4+b^4+c^4=9$

- Jul 26th 2007, 07:55 AM #4