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If a+b+c=3 then what is the maximum valuethe equation:a/(a^3+b^2+c) + b/(b^3+c^2+a) + c/(c^3+a^2+b)

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one month ago

NANDEESH
13 Points
```							16t²–40t–119=0Write 119 as a factor of its prime factors. 119=7*17 Now,See if we can find two numbers such that:Product=16*(–119)And,Sum=–40 Thus, the numbers are: 28 and–68 Therefore, the equation becomes:=〉16t²–68t+28t–119=0=〉 4t(4t–17)–7(4t–17)=0=〉(4t–7)(4t–17)=0 =〉t = 7/4 or 17/4Thus,7/4 and17/4 are the required solutions!
```
one month ago
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• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions