A+b+c=9and ab+bc+ca=23 find a²+b²+c²
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a³+b³+c³- 3abc = (a+b+c) (a²+b²+c²-ab-bc-ca)
a+b+c= 9 (given)
ab+ bc + ca = 23 (given)
⇒ 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 eqation 1 we get,
a³+b³+c³- 3abc
= 9 (35-23)
= 108
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