**(1)**Solve:-
\[\alpha^2+\alpha\beta+\beta^2=\gamma^2\]
Where \(\alpha,\beta,\gamma\) are positive distinct primes

**(2)**
Find out all roots of:-
\[x^4+9x^3+1\]

**(3)**
Find out the discriminant of:-

\[x^3+x^2+x+1\]

**(3)**
Solve the following differential equation:-
\[\frac{d}{dx}(y)=\frac{x^3+x^2y+xy^2+y^3}{x^3+x^2y+y^3}\]

**(4)**
Find out the sum of all values of \(x\) such that:-
\(x^4+x^3+x^2+x+1\) is a perfect square

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## Comments

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TopNewest3) \(ax^3 + bx^2 + cx +d\) has discriminant as \(b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd\)

Thus given discriminant is -16

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Is there any general definition for "Determinant" for n-degree equation?

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This Wiki might help you Thanks

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@Deepanshu Gupta too

Need some time , involvingLog in to reply

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\( \dfrac{dy}{dx} = \dfrac{1 + \dfrac{y}{x} + (\dfrac{y}{x})^2 + (\dfrac{y}{x})^3}{1 + \dfrac{y}{x} + (\dfrac{y}{x})^3}\)

\( \dfrac{y}{x}= v\)

\( x\dfrac{dv}{dx} + v = \dfrac{1 + v + v^2 + v^3}{v^3 + v + 1}\)

\( x\dfrac{dv}{dx} = \dfrac{1 - v^4}{v^3 + v + 1}\)

\( \dfrac{v^3 + v + 1}{1 - v^4} dv = \dfrac{1}{x}dx\)

\( \dfrac{v(v^2+1)}{(1 + v^2)(1- v^2)} + \dfrac{1}{(1 + v^2)(1- v^2)} = \dfrac{1}{x}dx\)

\( \dfrac{v}{1 + v^2} + \dfrac{1}{2}( \dfrac{1}{1 + v^2} + \dfrac{1}{1 - v^2}) = \dfrac{1}{x}dx\)

=\( \dfrac{1}{2}ln{(1+v^2)} + \dfrac{1}{2}(tan^{-1}v) + \dfrac{1}{4}(ln(\dfrac{1 - v}{1 + v}) = lnx + lnC\)

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4) \(x^5 = 1\)

\( \dfrac{x^4}{x^5} + \dfrac{x^3}{x^5} + x^2 + x + 1 =0\)

\( x^2 + \dfrac{1}{x^2} + x + \dfrac{1}{x} + 1 =0\)

\( Keep~ y = x + \dfrac{1}{x}\)

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How did you get \( x^5 = 1 \)?

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You used the Fifth root of Unity here ? Right?

Where \(w^5 = 1\)

However, how will this give all possible values of

xfor which the given expression is a perfect square.Log in to reply

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You know this - sum of GP

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But why you are finding root???? I mean how..will this give all values of x such that the given expression is perfect square.......i am not able to understand your solution..plz help

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1) Please comment on this,

Rhs can't be even as sum of 3 odd primes will give always a prime,

If \(\alpha =2 , Then~ \beta\) will be an odd prime, and on RHS is also odd prime ,thus \(\alpha=2~or\beta=2\)

\( 4 + 2\alpha + \alpha^2 = \gamma^2\)

\( (\alpha + 1)^2 - \gamma^2 = -3\)

\( This ~can't~be ~possible~for~primes \to \boxed{(\alpha + \gamma + 1)(\alpha - \gamma + 1) = -3}\)

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Hint:-complete the square by adding \(\alpha\beta\) both sides

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For 2, plot \(y=x^4+9x^3\) and \(y=-1\) and check the points of intersection.

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Very nice thinking

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Thanks!

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For (1).. 3,5,7 satisfies the condition.... @megh choksi sum of odd primes is not always prime as evident.....

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You are right with 3, 5 and 7. I would like to answer to only question that I feel more interesting.

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2) For x^4 + 9 x^3 + 1 = 0 {Please write an equation.}

-0.489801491372372

-8.998627630183605

0.244214560777989 + j 0.408953683836186

0.244214560777989 - j 0.408953683836186

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Megh Choksi--- I think there is some mistake between the 3rd and 4th step!! as there is a term remaining in numerator that is v^3

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Thanks , so the expression becomes \(x\dfrac{dv}{dx} = \dfrac{1 + v^3 - v^4}{1 + v^3 - v^4}\)

\( \dfrac{v^3 + v + 1}{1 + v^3 - v^4} dv = \dfrac{1}{x}dx\)

\( 1 + \dfrac{v + v^4}{1 + v^3 - v^4} = \dfrac{1}{x}dx\)

then its simple

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Am I right? let \(\alpha\)=a \(\beta\)=b \(\gamma\)=c Then, \(a^2+ab+b^2-c^2\)=o a=\(\frac{-b \pm \sqrt{b^2-4b^2+4c^2}}{2}\)...........(1) The discriminant should be a perfect square.\(\Rightarrow 4c^2-3b^2\)=0 or a perfect square. \(\Rightarrow\) c=b= prime perfect square.(couldn't there be any other possibilities where b is not equal to c ?please check ) on substituting in (1) we can get values for \(\alpha\) , \(\beta\), \(\gamma\).

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discriminant -16??

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