Question

$\begin{array}{c}\text{If}a+b+c=5\text{and}ab+bc+ca=10\text{,}\text{then prove that}{a}^3+b^3+c^3-3abc=-25.\end{array}$

Answer 1

Given that, $a+b+c=5$ $ab+bc+ca=10$

$(a+b+c{)}^3=a^3+b^3+c^3+3(a+b+c\left)\right(ab+ac+bc)-3abc$

$=5^3=a^3+b^3+c^3+\mathrm{3.5.10}-3abc$

$a^3+b^3+c^3-3abc=125-150=-25$