Question

if (a-b)=3, (b-c)=5 and (c-a)=1 then the value of (a³+b³+c³-3abc)÷a+b+c is A) 17.5. B) 20.5 C)10.5. D)15.5

Answer 1

The value of (a³+b³+c³-3abc)÷a+b+c is 17.5

Explanation:

Using the identity:

a^3+b^3+c^3 – 3 abc = (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

= 17.5

Hence the value of (a³+b³+c³-3abc)÷a+b+c is 17.5

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