This question was uploaded on 12/10/21 on social media accounts.
Three real numbers a, b, c such that their sum is 7, the sum of their cubes is 73, their product is 8. Find the sum of their reciprocals.
Solution:
Given equations:
\[\color{blue} {abc=8}\]
\[\color{red} {a+b+c=7}\]
\[\color{green} {a^3+b^3+c^3=73}\]
\[\color{purple}{\frac1a+\frac1b+\frac1c=x}\]
Now:
\[\color{fuchsia} {a^2+b^2+c^2}=(\color{red}{a+b+c})^2\]
\[-2\color{blue} {abc}\left(\color{purple}{\frac1a+\frac1b+\frac1c}\right)\]
\[\Rightarrow\color{fuchsia} {a^2+b^2+c^2}=\color{red}{49}-\color{blue} {16}\left(\color{purple}{x}\right)--(1)\]
Now:
\[\color{green} {a^3+b^3+c^3}-3\color{blue} {abc}=\]
\[(\color{red} {a+b+c})\left(\color{fuchsia} {a^2+b^2+c^2}-\color{blue} {abc}\left(\color{purple}{\frac1a+\frac1b+\frac1c}\right)\right)\]
\[\Rightarrow\color{green} {73}-\color{blue} {24}=(\color{red} {7})\left(\color{fuchsia} {a^2+b^2+c^2}-\color{blue} {8}\left(\color{purple}{x}\right)\right)----(2)\]
From (1) and (2):
\[x = \frac1a+\frac1b+\frac1c = \frac74\]
Also:
{a, b, c} = {1, 2, 4}
Puzzle related to Algebra, Equation.
