## Question

**Solve the following equations and find the value of abc?**

**If a + b + c = 6, a**^{2} + b^{2} + c^{2} = 14, and a^{3} + b^{3} + c^{3} = 36, then abc is,

^{2}+ b

^{2}+ c

^{2}= 14, and a

^{3}+ b

^{3}+ c

^{3}= 36, then abc is,

A. 12

B. 9

C. 6

D. 3

**Answer: C. 6**

**Solution: **

**From the given equations,**

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

=> (6)^{2} = 14 + 2(ab + bc + ca)

=> 36 – 14 = 2(ab + bc + ca)

=> 22/2 = ab + bc + ca

=> ab + bc + ca = 11

**a ^{3} + b^{3} + c^{3} − 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} − ab − bc − ca)**

= (6) (14 – 11)

= (6)(3)

= 18

Since, a^{3} + b^{3} + c^{3} − 3abc = 18

36 – 3abc = 18

-3abc = 18 – 36

-3abc = -18

**abc = 6.**

**Read more: **

**If a^2 + b^2 + c^2 = 20 and a + b + c = 0, find ab + bc + ca.**

**If a + b = 12 , b + c = 17 and c + a = 11 what is the value of a + b + c?**