Question

What is the value of a3+b3+c3-3abc/(a-b)2+(b-c)2+(c-a)2?

Answer 1

Answer:

it is zero

Step-by-step explanation:

in numerator. take 3 common

in denominator take 2 common

we get

3(a+b+c- abc)/2((a-b)+(b-c)+(c-a))

3(a+b+c- abc)/2(a-b+b-c+c- a)3(a+b+

in denominator all positive and negative values get cancel

we get

(a+b+c-abc)/2(0)

something multiplied by zero is zero

a+b+c-abc/0

something divided by 0 is 0

so answer is 0