If a+b+c= 7 and ab + bc + ca = 21, find the value of a³ + b³ + c³-3abc
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Step-by-step explanation:
a+b+c=7
Squaring both sides
(a+b+c)² = 7²
=a² + b²+ c² +2(ab+bc+ca)= 49. [ place value of ab+ = a² + b² + c² + 2(21) =49 bc+ca=21]
= a² + b² + c² + 42 =49
= a²+b²+c²= 49-42
= a²+b²+c²= 7
Now,
a³ + b³ + c³- 3abc = (a+b+c) (a²+ b²+ c² -ab -bc -ca)
= (7) [7-(21)]
= (7) (-14)
= -98
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