if a+b+c = 9 and ab+bc+ca = 23 then a cube + b cube + c cube- 3abc=

a³+b³+c³- 3abc = (a+b+c) (a²+b²+c²-ab-bc-ca)

a+b+c= 9 (given)

ab+ bc + ca = 23 (given)

So, a³+b³+c³- 3abc = 9 (a²+b²+c²- 23)            ... (1)

 

(a+b+c)²= a²+b²+c²+ 2ab+ 2bc+ 2ca

9²= a²+b²+c²+ 2(ab+bc+ca)

81= a²+b²+c²+ 2(23)

81 = a²+b²+c²+ 46

a²+b²+c²= 35

 

From (1), we get,

a³+b³+c³- 3abc = 9 (35-23) = 108