Question

If a+b+c= 12 and a³+b³+c³= 90.
Find a³+b³+c³-3abc

Please answer it fast....

Answer 1

Given (a + b + c) = 9
Squaring on both the sides, we get
(a + b + c)2 = 92
⇒ a2 + b2 + c2  + 2(ab + bc + ca) = 81
⇒ 35  + 2(ab + bc + ca) = 81
⇒ 2(ab + bc + ca) = 81 – 35 = 46
⇒ ab + bc + ca = 23  →  (1)
Recall that a3+b3+c3-3abc = (a + b + c)( a2 + b2 + c2 – ab – bc – ca)
                                       = 9(35 – 23)
                                       = 9(12) = 108

Answer 2

$\huge\mathfrak{Question}$
$<b><i>$
If a+b+c= 12 and a³+b³+c³= 90.Find a³+b³+c³-3abc?

$\huge\mathfrak{Answer}$

a³+b³+c³-3abc=(a+b+c) (a²+b²+c²-ab-bc-ac)
Given a+b+c=12,a²+b²+c²=90

(a+b+c)²=(a²+b²+c²)+2(ab+bc+ac)
12²=90+2×(ab+bc+ac)
54=2(ab+bc+ac)
(ab+bc+ac)=54/2=27

a³+b³+c³-3abc=12×90-(27)
                    =12×63
                    =756

#Regards...❤️