2021-12-09

Four system of equations on first, second, third, fourth degrees

This question was uploaded on 08/12/21 on social media accounts.



Solution:

Given equations:
\[\color{blue} {a+b+c = 4}\]
\[\color{fuchsia} {a^2+b^2+c^2 = 10}\]
\[\color{green} {a^3+b^3+c^3 = 22}\]
Now,
\[2(\color{maroon} {ab+bc+ca}) = (\color{blue} {a+b+c}) - (\color{fuchsia} {a^2+b^2+c^2})\]
\[\Rightarrow \color{maroon} {ab+bc+ca = 3}\]
\[3(\color{orange} {abc}) = (\color{green} {a^3+b^3+c^3}) - (\color{blue} {a+b+c})[(\color{fuchsia} {a^2+b^2+c^2}) - (\color{maroon} {ab+bc+ca})]\]
\[\Rightarrow \color{orange} {abc = -2}\]
\[\color{purple} {a^2b^2+b^2c^2+c^2a^2} = (\color{maroon} {ab+bc+ca})^2 - 2(\color{orange} {abc})(\color{blue} {a+b+c})\]
\[\Rightarrow \color{purple} {a^2b^2+b^2c^2+c^2a^2 = 25}\]
Finally:
\[\color{red} {a^4+b^4+c^4} = (\color{fuchsia} {a^2+b^2+c^2})^2 - 2(\color{purple} {a^2b^2+b^2c^2+c^2a^2})\]
\[\Rightarrow \color{red} {a^4+b^4+c^4 = 50}\]
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