Math, asked by ylniarb7997, 1 year ago

If a+b+c=6 and ab+bc+ca=1 find the value of a^3+b^3+c^3-3abc

Answers

Answered by Anonymous

3

a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-(ab+bc+ca))
6((34-1)
6×33
198

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