Can you simplify this expression?

Let

q = 108*a^3*(1+a^2)*(a^4-a^2+1) + 12*(1+3*a^6)*√(9*a^6 + 12).

This can be simplified as

q = 12*(9*a^3*(1+a^6) + (1+3*a^6)*√(9*a^6+12))

and using substitution for

b = a^3

further to

q = 12*(9*b*(1+b^2) + (1+3b^2)*√(9b^2+12)).

Using q, the expression can be written as

A = (q^(2/3) – 12) / (6q^(1/3)) = 1/6*q^(1/3) – 2*q^(-1/3).

As there are some cube roots, let’s try if A^3 gives anything simple:

A^3 = 1/216*q – 1/6*q^(1/3) + 2*q^(-1/3) – 8*q^(-1)

We can find two terms to match A exactly, thus

A^3 + A = q/216 – 8/q.

Let’s simplify 1/q in order to get rid of a √ in the denominator:

q = 12*(9*b*(1+b^2) + (1+3b^2)*√(9b^2+12)).

1/q = 1/12 * 1/(9*b*(1+b^2) + (1+3b^2)*√(9b^2+12)) =

= 1/12 * (9*b*(1+b^2) – (1+3b^2)*√(9b^2+12)) / ((9*b*(1+b^2) + (1+3b^2)*√(9b^2+12)) * (9*b*(1+b^2) – (1+3b^2)*√(9b^2+12))) =

= 1/12 * (9*b*(1+b^2) – (1+3b^2)*√(9b^2+12)) / (81*b^2*(1+b^2)^2 – (1+3b^2)^2*(9b^2+12)) =

= (9*b*(1+b^2) – (1+3b^2)*√(9b^2+12)) / 12 * 1/(-12) =

= -(9*b*(1+b^2) – (1+3b^2)*√(9b^2+12)) / 144.

Plugging this into

A^3 + A = q/216 – 8/q,

we find the right hand side to be equal to

(9*b*(1+b^2) + (1+3b^2)*√(9b^2+12)) / 18 + (9*b*(1+b^2) – (1+3b^2)*√(9b^2+12)) / 18 = b*(1+b^2),

so that now we have an equation

A*(1+A^2) = b*(1+b^2)

and as f(x) = x*(1+x^2) is monomorphic on R,

A = b = a^3.

Conclusion: the expression can be simplified to a^3 for all a in R.

151

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