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If (a-b)=3 (b-c)=5 and (c-a)=1 then find the value of (a^3+b^3+c^3-3ab)÷a+b+c

If (a-b)=3 (b-c)=5 and (c-a)=1 then find the value of (a^3+b^3+c^3-3ab)÷a+b+c

Grade:9

2 Answers

Vikas TU
14149 Points
6 years ago
Dear Student,
using identity
        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)÷a+b+c = (2/2)( a^2+b^2+c^2-2ab-2bc-2ac)
= (1/2){ (a – b)^2 + (b – c )^2 + (c – a)^2 }
= (1/2) { 9 + 25 + 1 } = 35/2 .
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)
Harsh Vardhan Sharma
25 Points
6 years ago
We studied about an identitya^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)Looking at the identity, we can conclude:-(a^3+b^3+c^3-3abc)÷a+b+c = (2/2)( a^2+b^2+c^2-2ab-2bc-2ac)= (1/2){ (a – b)^2 + (b – c )^2 + (c – a)^2 }= (1/2) { 9 + 25 + 1 } = 35/2 = 17.5Best of luck for your exams.

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