[ Completing the square. Sums of squares ]

In the expression $x^{6}+x^{4}+x A$, replace $A$ with a monomial so that a perfect square is obtained. Find as many solutions as possible

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Since the given expression is a trinomial, it will become a perfect square in one of three cases.

1) $x^{6}$ and $x^{4}$ are squares of two numbers, and $x A$ is twice the product of these numbers. We get $x^{6}+x^{4} \pm 2 x^{5}=\left(x^{3} \pm\right.$ $\left.x^{2}\right)^{2}$, that is, $A= \pm 2 x^{4}$.
2) $x^{4}$ and $x A$ are squares of two numbers, and $x^{6}$ is twice the product of these numbers. We get $x^{6}+x^{4}+1 / 4 x^{8}=\left(x^{2}+1 / 2\right.$ $\left.x^{4}\right)^{2}$, that is, $A=1 / 4 x^{7}$.
3) $x^{6}$ and $x A$ are squares of two numbers, and $x^{4}$ is twice the product of these numbers. We get $x^{6}+x^{4}+1 / 4 x^{2}=\left(x^{3}+1 / 2\right.$ $x)^{2}$, that is, $A=1 / 4 x$.

## Answer

$A= \pm 2 x^{4}, 1 / 2 x^{7}, 1 / 4 x$