Question

*
x^a= 3 ,x^b=2 , x^c= 1/6
then a3 + b3 + c3 = ?

Answer 1

Step-by-step explanation:

x^a * x^b * x^c = 3*2*1/6

x^(a + b + c) = 1

Taking log both side

(a + b + c) logx = log1

(a + b + c) logx = 0

(a + b + c) = 0

So , a³ + b³ + c³ = 3abc