Question

if x3 -6x2 -6x +8 is a cubic polynomial and A,B,C are zeroes of polynomial then find the value of

[ (A2B2C2 ) + (AB +BC+AC) +( A+B+C) ]​

Answer 1

Correction : a² + b² + c² - ab - bc - ca.

= ( a² + b² + c² - ab - bc - ca )

By multiplying it by 2 and dividing it by 2.

= 2 ( a² + b² + c² - ab - bc - ca ) ÷ 2

= ( 2a² + 2b² + 2c² - 2ab - 2bc - 2ca ) ÷ 2

= ( a² + a² + b² + b² + c² + c² - 2ab - 2bc - 2ca ) ÷ 2

= ( a² + b² - 2ab + b² + c² - 2bc + a² + c² - 2ca ) ÷ 2

= [ ( a - b )² + ( b - c )² + ( a - c )² ] ÷ 2

Now , whatever is the value of  ( a - b ) , ( b - c ) and ( a - c ) but its square will be always positive, and 2 is also a positive number.

So, the sum of ( a - b )², ( b - c )² and ( a - c )² will be a positive number and if a positive number is divided by a positive number then the result is also a positive number.

So, it is a positive number,hence it can't be a negative number.

Proved.

                                                         OR

A^2+b^2+c^2-ab-bc-ac

=1/2(2a^2+2b^2+2c^2-2ab-2bc-2ac)

=1 /2{(a-b)^2+(b-c)^2+(c-a)^2}

if a,b,c are real numbers then,(a-b)^2,(b-c)^2and(c-a)^2 are positive square of every real number is positive. So,

a^2+b^2+c^2-ab-bc-ca is always positive

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