Question -
Answer -
(i)┬а(a тАУ c)2┬а=4 (a тАУ b) (b тАУ c)
Let us expand theabove expression
a2┬а+ c2┬атАУ2ac = 4(ab тАУ ac тАУ b2┬а+ bc)
a2┬а+4c2b2┬а+ 2ac тАУ 4ab тАУ 4bc = 0
(a + c тАУ 2b)2┬а=0
a + c тАУ 2b = 0
Since a, b, c are inAP
b тАУ a = c тАУ b
a + c тАУ 2b = 0
a + c = 2b
Hence, (a тАУ c)2┬а=4 (a тАУ b) (b тАУ c)
(ii)┬аa2┬а+ c2┬а+4ac = 2 (ab + bc + ca)
Let us expand theabove expression
a2┬а+ c2┬а+4ac = 2 (ab + bc + ca)
a2┬а+ c2┬а+2ac тАУ 2ab тАУ 2bc = 0
(a + c тАУ b)2┬атАУb2┬а= 0
a + c тАУ b = b
a + c тАУ 2b = 0
2b = a+c
b = (a+c)/2
Since a, b, c are inAP
b тАУ a = c тАУ b
b = (a+c)/2
Hence, a2┬а+c2┬а+ 4ac = 2 (ab + bc + ca)
(iii)┬аa3┬а+ c3┬а+6abc = 8b3
Let us expand theabove expression
a3┬а+ c3┬а+6abc = 8b3
a3┬а+ c3┬атАУ(2b)3┬а+ 6abc = 0
a3┬а+(-2b)3┬а+ c3┬а+ 3a(-2b)c = 0
Since, if a + b + c =0, a3┬а+ b3┬а+ c3┬а= 3abc
(a тАУ 2b + c)3┬а=0
a тАУ 2b + c = 0
a + c = 2b
b = (a+c)/2
Since a, b, c are inAP
a тАУ b = c тАУ b
b = (a+c)/2
Hence, a3┬а+c3┬а+ 6abc = 8b3