Squares, Then Cubes, Now Squared Squares

(This is not a recommended approach to the problem...)

"Hmm...I wonder if they are integers..."

"Well, what about 1, 2, and 3?"

"Oh, hmm, those work. Cool."

$1^4 +2^4 + 3^4 = 1 + 16 + 81 = 98.$

$a^2 +b^2+c^2=(a+b+c)^2-2(ab+ac+bc)=(6)^2-2(ab+ac+bc)=14\rightarrow\color{#20A900}{\boxed{(ab+ac+bc)=11}}$ $a^3+b^3+c^3=(a+b+c)^3-3[(a+b+c)(ab+ac+bc)-abc]=(6)^3-3[(6)(11)-abc]=36\rightarrow\color{#69047E}{\boxed{(abc)=6}}$ $a^4+b^4+c^4=(a+b+c)^4-12(abc)(a+b+c)-4[a^3(c+b)+b^3(c+a)+c^3(a+b)]-6[(ab+ac+bc)^2-2abc(a+b+c)]$ $a^4+b^4+c^4=(6)^4-12(6)(6)-4[a^3(6-a)+b^3(6-b)+c^3(6-c)]-6[(11)^2-2(6)(6)]$ $a^4+b^4+c^4=1296-432-4[(6a^3-a^4+6b^3-b^4+6c^3-c^4)]-294=$ $a^4+b^4+c^4=1296-432-4[6(a^3+b^3+c^3)-(a^4+b^4+c^4)]-294=$ $a^4+b^4+c^4=1296-432-4[6(36)-(a^4+b^4+c^4)]-294=$ $a^4+b^4+c^4=1296-432-864+4(a^4+b^4+c^4)-294=$ $a^4+b^4+c^4=4(a^4+b^4+c^4)-294=$ $\rightarrow\color{#D61F06}{(a^4+b^4+c^4)}=\frac{294}{3}=\Large\color{#3D99F6}{\boxed{98}}$

It can be guessed pretty easily

I prefer to use Vieta's formula and Newton's sums to do this.

From Vieta's formula, we have:

$a + b + c = 6$

$ab + ac + bc = x$

$abc = y$

To find $a^4+b^4+c^4$ , we will need to find $x$ and $y$ first.

Using Newton's sums:

$a^2+b^2+c^2 = (a+b+c)(a+b+c) - 2(ab + ac + bc)$

$14 = 6(6) - 2x\implies x=11\implies ab+ac+bc=11$

Next,

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

$36 = 6(14) - 11(6) -3y\implies y=-6 \implies abc=-6$

Then, substitute in here to find the answer:

$a^4+b^4+c^4 = (a+b+c)(a^3+b^3+c^3) - (ab+ac+bc)(a^2+b^2+c^2) -abc(a+b+c)\\ =6(36) - 11(14) -(-6)(6) = \boxed{98}$

Note: Is this really a level 1 Algebra question?

1+2+3 = 6 and they cannot be the same number as 2+2+2 so I used this method 1+2+3 and squared it , cubed it and I did it to the power of 4

BTW i am only 10 yrs old