This question was uploaded on 26/12/21 on social media accounts.
Solution:
Given equations:
\[\color{red} {a^2b+ab^2 = ab(a+b) = 1}---(1)\]
\[\color{blue} {a^3+b^3 = (a+b)(a^2-ab+b^2) = 2}---(2)\]
\[\color{green} {a^3-b^3 = (a-b)(a^2+ab+b^2) = ?}\]
2(1) + (2):
\[\color{red} {2ab(a+b)}+\color{blue} {(a+b)(a^2-ab+b^2)} = 2(1)+2\]
\[\Rightarrow \color{fuchsia} {(a+b)(a^2+ab+b^2) = 4}---(3)\]
3(1) + (2):
\[\color{red} {3ab(a+b)}+\color{blue} {(a+b)(a^2-ab+b^2)} = 3(1)+2\]
\[\Rightarrow \color{orange} {(a+b)(a^2+2ab+b^2) = 5}\]
\[\Rightarrow\color{orange} {(a+b)(a+b)^2 = 5}\]
\[\Rightarrow \color{orange} {(a+b)^3 = 5}---(4)\]
(2) - (1):
\[\color{blue} {(a+b)(a^2-ab+b^2)} + \color{red} {ab(a+b)} = 2 - 1\]
\[\Rightarrow \color{purple} {(a+b)(a^2-2ab+b^2) = 1}\]
\[\Rightarrow \color{purple} {(a+b)(a-b)^2 = 1}---(5)\]
(3)2(5):
\[\color{fuchsia} {(a+b)^2(a^2+ab+b^2)^2}\color{purple} {(a+b)(a^2-2ab+b^2)} = (4)^2(1)\]
\[\left(\color{green} {(a-b)(a^2+ab+b^2)}\right)^2 \left(\color{orange} {(a+b)}\right)^3 = 16\]
From equation (4):
\[\left(\color{green} {a^3-b^3}\right)^2 \left(\color{orange} {5}\right) = 16\]
\[\left(\color{green} {a^3-b^3}\right)^2 = \frac{16}5\]
\[\color{green} {a^3-b^3} = \pm\frac4{\sqrt5}\]
