Question

If a+b+c=3, a^2+b^2+c^2=6 and 1/a+1/b+1/c=1 where a,b,c are all non-zero, then 'abc' is equal to?

Answer 1

Answer:

3/2

Step-by-step explanation:

a +b +c = 3

squaring both sides

(a+b+c)^2 = 3^2

((a+b)+c)^2 = 3^2

(a+b)^2 + c^2 + 2(a+b)c = 9

a^2 + b^2 +2ab + c^2 + 2ac + 2bc = 9

a^2 + b^2 + c^2 +2ab + 2bc + 2ac= 9

a^2 + b^2 + c^2 = 6

6 +2ab + 2bc + 2ac= 9

2 (ab + bc + ac) = 3

(ab + bc + ac) = 3/2    eq1

1/a +1/b + 1/c = 1

multiplying by abc

bc + ac + ab = abc

abc = ab + bc + ac

ab + bc + ac = 3/2 from eq1

so abc = 3/2

Answer 2

Answer:

3/2 is the right answer

Step-by-step explanation:

jai hind