42. If a+b+c ≠ 0 ,then prove that a3+b3+c3=3abc,only when a = b = c.

Open in App

Solution

Suggest Corrections

Similar questions

a + b + c = 0

a^{3} + b^{3} + c^{3} = 3 a b c

a + b + c = 5

a b + b c + c a = 10

a^{3} + b^{3} + c^{3} - 3 a b c = - 25

> 0

\frac{a^{3}}{b^{3}} + \frac{b^{3}}{c^{3}} + \frac{c^{3}}{a^{3}} \geq 3

a^{3} + b^{3} + c^{3} - 3 a b c

a + b + c = 0

If a+b+c=0 then $(a^{3} + b^{3} + c^{3})$ is
(a) 0 (b) abc (c) 2abc (d) 3abc

Join BYJU'S Learning Program