Problem 1. Find the maximum value of the expression $a^{3} b-b^{3} a$, if $a^{2}+b^{2}=1$.

Answer. $\frac{1}{4}$

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# Solution.

1st method.

Since $\mathrm{a}^{2}+\mathrm{b}^{2}=1$, there exists a number $\alpha$ such that $\mathrm{a}=\cos \alpha, \mathrm{b}=\sin \alpha$. Then $\mathrm{a}^{3} \mathrm{~b}-\mathrm{b}^{3} \mathrm{a}=\mathrm{ab}\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)$ $=\cos \alpha \sin \alpha\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right)=\frac{1}{2} \sin 2 \alpha \cos 2 \alpha=\frac{1}{4} \sin 4 \alpha$. The maximum value of this expression is $\frac{1}{4}$ and is achieved, for example, at $\alpha=\frac{\pi}{8}$.

## 2nd method.

Let $\mathrm{C}=\mathrm{a}^{3} \mathrm{~b}-\mathrm{b}^{3} \mathrm{a}=\mathrm{ab}\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)$. The square of this expression is $\mathrm{C}^{2}=\mathrm{a}^{2} \mathrm{~b}^{2}\left(\mathrm{a}^{4}+\mathrm{b}^{4}-2 \mathrm{a}^{2} \mathrm{~b}^{2}\right)=\mathrm{a}^{2} \mathrm{~b}^{2}$ $\left(\left(a^{2}+b^{2}\right)^{2}-4 a^{2} b^{2}\right)=a^{2} b^{2}\left(1-4 a^{2} b^{2}\right)$.

Let $\mathrm{x}=4 \mathrm{a}^{2} \mathrm{~b}^{2}$. Then $C^{2}=\frac{x(x-1)}{4}$. The maximum value of the expression $\mathrm{x}(1-\mathrm{x})$ is $\frac{1}{4}$ and is achieved at $\mathrm{x}=\frac{1}{2}$. Then the maximum value of $\mathrm{C}^{2}$ is $\frac{1}{16}$. Note that the set of values of C is symmetric about zero, since if the values of a and b are swapped, the value of the expression changes to the opposite. Therefore, the maximum value of C is $\frac{1}{4}$.