Question

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

Answer 1

proof:

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac$
$(5)^2=a^2+b^2+c^2+2(ac+bc+ab)$
$25=a^2+b^2+c^2+2(10)$
[tex]25=a^2+b^2+c^2+20 [/tex]
∴$5=a^2+b^2+c^2$.....(i)

$a^3 + b^3 + c^3 - 3abc =(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$
$a^3+b^3+c^3-3abc=(5)(5-(ab+bc+ac))$
$a^3+b^3+c^3-3abc=(5)(5-10)=(5)(-5)$
∴$a^3+b^3+c^3-3abc=-25$   Hence proved
                                     Hope It Helps

Answer 2

Answer:

Step-by-step explanation: