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Question
If \[\frac{a}{b} + \frac{b}{a} = 1\] then a3 + b3 =
Options
- 1
-1
- \[\frac{1}{2}\]
0
Solution
Given `a/b + b/a = 1`
`(a xx a)/(b xx a) +(b xx b) /(a xx b) = 1`
`a^2/(ab) +b^2/(ab) = 1`
`(a^2 +b^2 ) /(ab )= 1`
`a^2 +b^2 = 1 xx ab`
`a^2 +b^2= ab`
`a^2 +b^2 - ab = 0`
Using identity `a^3 +b^3 = (a+b)(a^2 - ab +b^2)`we get,
`a^3 +b^3 = (a+b)(a^2 - ab + b^2)`
`a^3 +b^3 = (a+b)(0)`
`a^3 +b^3 = 0`
Hence the value of `a^3 + b^3 ` is 0 .
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