Question

if a+b+c=5 & ab+bc+ba=10 then find a³+b³+c³-3abc

Answer 1

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Answer 2

a+b+c=5

ab+bc+ca=10

Since

(a+b+c) ^2 = 25

a ^ 2 + b^2 +c^2 + 2(ab+bc+ca)=25

a ^ 2 +b ^ 2 +c ^2 + 2×10=2

a ^ 2 +b ^2 +c ^ 2 +2×10=25

a ^ 2 +b ^ 2 +c ^2 =5

We know that,

a ^ 3 +b ^ 3+c ^3 − 3abc=

(a+b+c) (a ^ 2 +b ^2 +c ^2 −(ab+bc+ca ))

a ^ 3 +b ^ 3 +c ^ 3 −3abc=(5)×(5−10)

a ^ 3+b ^ 3 +c ^ 3 −3abc=−25

Hence proved

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