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I'm having difficulty with the following question enter image description here

The first part seems simple enough, and by expanding the right hand side I get that
$p=-a^2+b+c$
$q=a(b-c)$
$r=bc$

But when asked to show that $a^2$ is a root of the mentioned equation, I'm unsure as to tackle the question. I tried substituting, $u$ with $a^2$ along with $p,q$ and $r$ to be in terms of $a,b$ and $c$. But that gives a sum with a lot of variables, and no clear way on how to conclude that this sum is equal to $0$. I see no other way to answer the question without substituting $p,q$ and $r$ and using some tricks to cancel out all the terms.

Any help on this question would be much appreciated.

Andrew Brick
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    This is the method that Descartes used to solve the quartic. It is a significant improvement over the original method of Ferrari-Cardano. – André Nicolas Jul 08 '15 at 03:31

2 Answers2

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Substituting for $p$, $q$, and $r$ in the cubic, we have$$u^3 + 2(b + c - a^2)u^2 + ((b + c - a^2)^2 - 4bc)u - (ab - ac)^2 = 0.$$This is equivalent to $$u^3 + 2(b + c - a^2)u^2 + (b^2 + c^2 - 2bc - 2a^2b - 2a^2c)u - a^2(b-c)^2 = 0.$$Now put $u = a^2$ to give$$a^6 + 2ba^4 + 2ca^4 - 2a^6 + b^2a^2 + c^2a^2 + a^6 - 2bca^2 - 2a^4b - 2a^4c - a^2b^2 + 2a^2bc - a^2c^2 = 0,$$which means that $u = a^2$ is a root of the cubic.

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Another way to carry out the algebra is to notice that the first three terms resemble a perfect square, and indeed we may write

$$u^3 + 2 p u^2 + (p^2-4r) u -q^2=u(u+p)^2-4r u-q^2.$$

But for $u=a^2$ we have $u+p=b+c$, so plugging in the substitutions gives $$a^2(b+c)^2-4bca^2-a^2(b-c)^2=a^2((b+c)^2-(b-c)^2-4bc)=0$$ upon simplifying the difference of squares. So $u=a^2$ is indeed a root.

Semiclassical
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