If a + b + c = 12, then a² + b² + c² = 90, then find the value of a³ + b³ + c³ - 3abc.
SUPER URGENT!!!!! PLEASE!!!!!
Adithya1234:
is tis crrct?
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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
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
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