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ax+by=3 ; ax^2+by^2=7 ; ax^3+by^3=16 ; ax^4+bx^4=42 ; Then find ax^5+bx^5 given a,b,x,y are real nos.

Note by Avinash Panneerselvam 4 years, 8 months ago

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Solution : Note that , for any \( n >0 , (ax^n+by^n)(x+y)-xy(ax^{n-1}+by^{n-1}) = ax^{n+1}+by^{n+1}\)

So now For \( n=2 \) we get \( 7(x+y)-3xy=16 \) and for \( n=3\) we get \( 16(x+y)-7xy=42 \) . Solving gives , \( x+y=-14 , xy=-38 \)

now again using the identity ,

\( ax^5+by^5 = (42)(-14)-(16)(-38) = \boxed{20} \)

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a/(ax - 1) + b/(bx - 1) = a + b . Find x .

Is it \(ax^4 + by^4 = 42, ax^5 + by^5\)? I think you typed wrongly.

If it is, then the answer should be \(20\).

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## Comments

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TopNewestSolution : Note that , for any \( n >0 , (ax^n+by^n)(x+y)-xy(ax^{n-1}+by^{n-1}) = ax^{n+1}+by^{n+1}\)

So now For \( n=2 \) we get \( 7(x+y)-3xy=16 \) and for \( n=3\) we get \( 16(x+y)-7xy=42 \) .

Solving gives , \( x+y=-14 , xy=-38 \)

now again using the identity ,

\( ax^5+by^5 = (42)(-14)-(16)(-38) = \boxed{20} \)

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a/(ax - 1) + b/(bx - 1) = a + b . Find x .

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Is it \(ax^4 + by^4 = 42, ax^5 + by^5\)? I think you typed wrongly.

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If it is, then the answer should be \(20\).

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