100 marks
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✂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
Answers
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)
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!!!