Algebra and Applications Test 3

Question : 1 of 25

Marks: +1, -0

a+b+c=6

a^2+b^2+c^2=14

and

a^3 + b^3 + c ^3 = 36

, then the value of

a b c

is:

Solution:

a + b + c = 6

a^2 + b^2 + c^2 = 14

a^3 + b^3 + c^3 = 36

(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

(6)^2 = (14) + 2(ab + bc + ca)

ab + bc + ca = 11

Now

a^3 + b^3 + c^3 - 3abc = (a + b + c) (a^2 + b^2+ c^2 - ab - bc - ca)

36 - 3abc = 6 (14 - 11)

36 - 3abc = 18

⇒ 3abc = 18 - 36

⇒abc=6

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