Question

100 marks
HURRY UP

✂For a = 2/3 , b= 4/5 ,c =-5/6, verify that ,
(a+b)+c=a+(b+c)
(ab)c=a(bc)
a(b-c)=ab-ac
​

Answer 1

a + b + c = 0 so c = -(a + b).

a^5 + b^5 + c^5

= a^5 + b^5 - (a + b)^5..... (1)

By binomial theorem,

(a + b)^5 = a^5 + 5a^4*b + 10a^3*b^2 + 10a^2*b^3 + 5ab^4 + b^5.

Putting it in (1),

a^5 + b^5 + c^5

=-5ab (a^3 + 2a^2*b + 2ab^2 + b^3).

= -5ab (a^3 + b^3 + 2a^2*b + 2ab^2)

= -5ab [ (a + b)(a^2 -ab + b^2) + 2ab (a + b) ].

= -5ab (a + b) (a^2 + ab + b^2)....(2)

ab + bc + ca

= ab + c (a + b).

= ab - (a + b)^2 - from (1)

= -(a^2 + ab + b^2)...... (3)

From (2) & (3),

(a^5 + b^5 + c^5)/(ab + bc + ca)

= 5ab (a + b)

= -5abc - from (1)

Answer 2

For a = 2/3, b= 4/5, c =-5/6

Put the values and then solve:

(a+b)+c = (2/3 + 4/5) + (-5/6)

= ((10 + 12)/15) - (5/6)

= (22/15) - (5/6)

= (44 - 25)/30 = 19/30

Similarly, find a+(b+c)= 19/30

and similarly solve all.

Now, Solving similarly we'll get:

a + (b + c) = (a + b) + c

(ab)c=a(bc)

a(b-c)=ab-ac

Thankyou!!!