Problem 1112

I. Solution: Known identity

$$\cos ^{2} x=1-\sin ^{2} x$$

Raising both sides to the third power

$$\cos ^{6} x=1-3 \sin ^{2} x+3 \sin ^{4} x-\sin ^{6} x$$

thus

$$\sin ^{6} x+\cos ^{6} x-1=3 \sin ^{2} x\left(1-\sin ^{2} x\right)=-3 \sin ^{2} x \cos ^{2} x$$

Raising both sides to the third power, we obtain the identity to be proven. This completes the proof of our theorem, as we started from a true identity and performed only unambiguous operations throughout.

Tibor Harza (Székesfehérvár, József Attila g. I.o.t.)

II. Solution: The left side is the sum of the cubes of two numbers, which can be factored into the product of two factors according to the identity $a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)$, one of which is the sum of the two numbers. It is sufficient to prove that the latter is equal to zero, that is, it is to be proven that

$$\left[\left(\sin ^{2} x\right)^{3}+\left(\cos ^{2} x\right)^{3}-1\right]+3 \sin ^{2} x \cos ^{2} x=0$$

Let $\sin ^{2} x=u$, then $\cos ^{2} x=1-u$, and thus the left side of (1) is

$$u^{3}+(1-u)^{3}-1+3 u(1-u)=u^{3}+1-3 u+3 u^{2}-u^{3}-1+3 u-3 u^{2}$$

which is indeed identically equal to 0, regardless of $u$.

Tibor Tolnai (Szombathely, Nagy Lajos g. II.o.t.)