Question

Solve $a+b+c=3$
$a^2+b^2+c^2=6$
$a^3+b^3+c^3=8$
then find $a^4+b^4+c^4=?$

Answer 1

$a+b+c=2$.......(A)

$a^2+b^2+c^2=6$.......(B)

$a^3+b^3+c^3=8$.......(C)

Now, take whole square of equation (A)

$(a+b+c)=2^2$

$\Rightarrow a^2+b^2+c^2+2ab+2bc+2bc+2ca=4$

$\Rightarrow 2(ab+bc+ca)=4-(a^2+b^2+c^2)$

$\Rightarrow 2\left(\right)=4-6$.....from (B)

$\Rightarrow 2(ab+bc+ca)-2$

$(ab+bc+ca)=-1$........(1)

Now,

$(a+b+c)(a^2+b^2+c^2)=a^3+b^3+c^3+a^{2}b+b^{2}a+b^{2}c+c^{2}b+c^{2}a+a^{2}c$

and

$(a+b+c)(ab+bc+ca)=a^{2}b+b^{2}a+b^{2}c+c^{2}a+c^{2}a+a^c+3abc$

$\Rightarrow (a^{2}b+b^{2}a+b^{2}c+c^{2}b+c^{2}a+a^{2}a+a^{2}c)=(a+b+c)(ab+bc+ca)-3abc$

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

$a^3+b^3+c^3-3abc=(a+b+c)\left[\right(a^2+b^2+c^2)-(ab+bc+ca\left)\right]$

Substitute values from equation A,B,C and 1.

$\Rightarrow 8-3abc=2\left[\right(6)-(-1\left)\right]$

$\Rightarrow 8-3abc=2\left(7\right)$

$\Rightarrow -3abc=14-8$

$\Rightarrow -3abc=6$

$\Rightarrow abc=-2$........(3)

Squaring equation (2)

$(a^2+b^2+c^2)=6^2$

$\Rightarrow (a^4+b^4+c^4)+2(a^2{b}^2+b^2{c}^2+c^2{a}^2)=36$

$\Rightarrow a^4+b^4+c^4=36.2(a^2{b}^2+b^2{c}^2+c^2{a}^2)$.......(4)

And

$(ab+bc+ca{)}^2=a^2{b}^2+b^2{c}^2+c^2{a}^2+2(a^{2}bc+a{b}^{2}c+ab{c}^2),so$

$\Rightarrow a^2{b}^2+b^2{c}^2+c^2{a}^2=(ab+bc+ca{)}^2-2abc(a+b+c)$

Substitute values from A, 1 & 3

$\Rightarrow a^2{b}^2+b^2{c}^2+c^2{a}^2=(-1{)}^2-2(-2\left)\right(2)$

$\Rightarrow a^2{b}^2+b^2{c}^2+c^2{a}^2=1+8$

$\Rightarrow a^2{b}^2+b^2{c}^2+c^2{a}^2=9$

Substitute the above value in equation (4)

$\Rightarrow a^4+b^4+c^4=36-2\left(9\right)$

$a^4+b^4+c^2=18$