Math, asked by srijana1453, 1 year ago

If a+b+c=4 and ab+bc+ca=6,find the value of a^3+b^3+c^3-3abc.

Answers

Answered by ymeena1848
1
(a+b+c)²=a²+b²+c²+2(ab+bc+ca)
=>4²=a²+b²+c²+2×6
=>a²+b²+c²=16-12
=>a²+b²+c²=4

a³+b³+c³-3abc= (a+b+c)[a²+b²+c²-(ab+bc+ca)]
                       =4(4-6)
                       =-8
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