Question

If 1/a+1/b+1/c= 1/a+b+c than prove that 1/a7+1/b7+1/c7= 1/a7+b7+ c7​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Step-by-step explanation:

1/a + 1/b + 1/c = 1/(a+b+c)

=> 1/a + 1/b = 1/(a+b+c) -1/c

=> (b+a)/ab = {c - (a+b+c)/c(a+b+c)

=> (a+b)/ab = -(a+b)/(ac+bc+c^2)

=> 1/ab = -1/(ac+bc+c^2)

=> -ab = ac+bc+c^2

=-c^2 = ab+bc+ca

Similarly we can show that

-b^2 =ab+bc+ca and -c^2 = ab+bc+ca

So a=b=c. since -a^2 = -b^2 = - c^2

Then 1/a + 1/b + 1/c = 1/(a+b+c) => 3/a= 1/3a

=> 1/a^5 + 1/b^5 + 1/c^5 =3/a^5=(( 3/a )(1/a^4)

= 1/3a × 1/a^4 = 1/3a^5 = 1/(a^5 +b^5+c^^5)

Hope it help