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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NANDEESH · Answer

16t²–40t–119 =0 Write 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 17/4 Thus,7/4 and17/4 are the required solutions!

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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)

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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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)

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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