Question

If a + b + c.= 5 and ab + bc +ca = 10 , then prove that a3+ b3+ c3-3abc = -25

Answer 1

Start with

$(a+b+c)^3=a^3+b^3+c^3+3(a^2b+a^2c+ab^2+ac^2+b^2c+bc^2)+6abc$

Given that $a+b+c=5$, we know $(a+b+c)^3=125$.

Also, $ab+bc+ca=10$, so

$a^2b+a^2c+ab^2+ac^2+b^2c+bc^2=a(ab+ac)+b(ab+bc)+c(ac+bc)$

$=a(10-bc)+b(10-ca)+c(10-ab)$

$=10(a+b+c)-3abc$

$=50-3abc$

So we have

$125=a^3+b^3+c^3+3(50-3abc)+6abc$

$125=a^3+b^3+c^3+150-9abc+6abc$

$\implies -25=a^3+b^3+c^3-3abc$

QED