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Four sets of equations on three variables with different powers. Find the sum of the fourth powers of all variables if sum their first, second, and third power is given.
Solution 1:
Let,
\[a\ge b\ge c\]
\[a+b+c=3----(1)\]
\[a^2+b^2+c^2=5----(2)\]
\[a^3+b^3+c^3=9----(3)\]
We know that:
\[a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)\]
\[\Rightarrow ab+bc+ca=2----(4)\]
Also,
\[a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\]
\[\Rightarrow abc=0----(5)\]
From above equation, at least one element is zero.
If a is zero then b,c are negative (Equation (1) doesn't satisfy).
If b is zero then a,c are positive, negative (Equation (4) doesn't satisfy).
c must be zero:
Now,
\[c=0\]
By equation (4),
\[ab=2\]
By equation (1),
\[a+b=3\]
\[\Rightarrow a=2,⠀b=1\]
\[\Rightarrow a^4+b^4+c^4=17\]
And, {a, b, c} = {0, 1, 2}
Solution 2: (General Method)
\[a+b+c=3----(1)\]
\[a^2+b^2+c^2=5----(2)\]
\[a^3+b^3+c^3=9----(3)\]
Square of Equation (2):
\[a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)=25----(4)\]
Square of Equation (1):
\[a^2+b^2+c^2+2(ab+bc+ca)=9\]
\[\Rightarrow ab+bc+ca=2----(5)\]
Square of Equation (5):
\[(a^2b^2+b^2c^2+c^2a^2)+abc(a+b+c)=4\]
\[\Rightarrow (a^2b^2+b^2c^2+c^2a^2)+3abc=4----(6)\]
Multiply (1) and (5):
\[a^2b+ab^2+b^2c+bc^2+c^2a+ca^2+3abc=6----(7)\]
Multiply (1) and (2):
\[a^3+b^3+c^3+a^2b+ab^2+b^2c+bc^2+c^2a+ca^2=15\]
\[\Rightarrow a^2b+ab^2+b^2c+bc^2+c^2a+ca^2=6----(8)\]
(8) in Equation (7):
\[3abc+6=0\]
\[\Rightarrow abc=0----(9)\]
(9) in Equation (6):
\[\Rightarrow a^2b^2+b^2c^2+c^2a^2=4----(10)\]
(10) in Equation (4):
\[a^4+b^4+c^4+2\times4=25\]
\[\Rightarrow a^4+b^4+c^4=17\]
Puzzle related to Algebra, Equation.
